JPJC 2026 Projectile Tutorial solutions

Kinematics Tutorial soln 2026/JPJC/PHYSICS/9478 1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS PROJECTILE MOTION TUTORIAL SOLUTIONS Self-Check Questions S1 The horizontal component of velocity. S2 The vertical component of velocity is zero, the acceleration = g = 9.81 ms-1 downwards. S3 All the kinematics quantities are independent of the mass. S4 When the resultant force on the object is zero. For a falling object it means that air resistance force pointing upwards is equal to the weight pointing downward. S5 Decreasing gradient magnitude, longer time to fall, lower speed when returned. Self-Practice Questions 1 Horizontal velocity and acceleration are constant. Vertical velocity is zero at highest point [ Ans: D ] 2 Option A has no effect. Options B will result in smaller range and smaller maximum height. Option D will result in a larger range and larger maximum height. 2 2 2 2 ( cos ) 12( sin ) / , using ( sin ) 0 2 2 cos sin / sin(2 ) / is maximum when =45. Other angles reduc es . Max height ( sin ) / (2 ). increases with y Range R u t t u g s u t gt R u g u g R R H u g H                  [ Ans: C ] 3 The time of flight will increase because the downward acceleration is lower. The range will be lower because the horizontal speed decreases continuously. [ Ans: A ] 4 The resultant force Y reduces to zero when the air resistance X increases to its maximum value, which is equal to the weight of the object. [ Ans: A ]

Kinematics Tutorial soln 2026/JPJC/PHYSICS/9478 2 5 The vertical initial velocity and acceleration are the same for X and Y. Horizontal velocity does not affect the time taken to travel the same vertical height. [ Ans: C ] Discussion Questions 1 (a) Take upward to be positive. 1 2 2 2 2 15sin(40 ) 9.64 m s 1 2 45 9.64 4.905 0 45 9.64 4.905 solving : 9.64 9.64 4(45)(4.905) 4.17 s, ignore '-' solution2(4.905) y y y u s u t gt t t t t t               (b)     15 cos(40 ) 15 cos(40 ) (4.17) 47.9 m xs t     (c) 1 2 2 2 2 1 15sin(40 ) 9.81(4.17) 31.27 m s [15cos(40 )] 31.27 33.3 m s y x y v v v v             tan 31.27 15cos(40 ) 2.72 69.8 below horizontal. y x o v v       (d) KE decreases until minimum and increases from minimum value until it hits the ground. The variation of GPE is opposite to that of KE. It increases until it reaches the maximum value at the highest point. It decreases thereafter. KE + GPE = constant.

Kinematics Tutorial soln 2026/JPJC/PHYSICS/9478 3 2 (a) vertical component: Take upward to be positive 1 2 2 2 2 3.0 2.3 0.7 m 6.5sin(50 ) 4.98 m s 1 2 0.7 4.98 4.905 4.905 4.98 0.7 0 4.98 4.98 4(4.905)(0.7) 0.169 s or 0.847 s2(4.905) taking the longer time, = 0.847 s y y s v s ut gt t t t t t t                  (The shorter time refers to the time when the ball is rising at 3.0 m) (b) Maximum height is reached when the vertical velocity is zero 2 2 2 2 0 [6.5sin(50)] 2(9.81) 1.26 m height 1.26 2.3 3.56 m y y y y y y v u a s s s         (c) Ball is moving with horizontal velocity at the top, vx=6.5cos(50 ) 2 2 1 2 1 (0.600)(6.5cos(50 ))2 5.24 J xKE mv    (d) Use the horizontal component to calculate the horizontal distance  6.5 cos(50) , 0.847 s from ( ) 3.54 m x x s t t a s    (e) Calculate the horizontal and vertical velocities separately, then combine them to calculate the final velocity. 1 1 2 2 2 2 1 6.5cos(50 ) 4.18 m s 6.5sin(50 ) 9.81(0.847) 3.33 m s (-ve sign means directed downward) 4.18 3.33 5.34 m s x y x y v v v v v               

Kinematics Tutorial soln 2026/JPJC/PHYSICS/9478 4 3.33angle given by tan 0.7966 4.18 38.5 below the horizontal. y x o v v       3(a) Assume that there is no air resistance and the kick is at an optimal angle of 450. 2 time of flight: 10 30(sin 45 ) (9.81)2 2(30sin 45 ) 9.81 4.32 s t t t      Horizontal distance = (30 cos45)4.32 = 91.6 m < 95 m. The initial velocity is not high enough. 3(b) Assume optimal angle of launch is 45° above horizontal. 2 time of flight: 10 9.5(sin 45 ) (9.81)2 2(9.5sin 45 ) 9.81 1.37 s t t t      Horizontal distance = (9.5 cos45 )1.37 = 9.2 m The world record is within the theoretical limit.  3(c) Let the initial velocity be u and time of flight = t. Take upward to be positive. 2 2 1 horizontal: 24.77 cos(38 ) ----- (1) 1vertical: 2.10 sin(38 ) (9.81) ----- (2)2 Substitute from (1) into (2) 9.81 24.772.10 = 24.77 tan(38 ) ( ) 2 cos38 Solving for , = 15.0 m s u t u t t t u u u           

Kinematics Tutorial soln 2026/JPJC/PHYSICS/9478 5 4(a) Air resistance always opposes motion of ball. (i) With air resistance, there are now two forces decelerating the ball. The ball hence experiences greater deceleration. As the deceleration is greater, the time taken for velocity to decrease to zero, at maximum height, is shorter. (ii) Since both Earth’s gravitational pull and air resistance are pointing downwards, this causes the object’s velocity to decrease at a faster rate. The maximum height reached by the object is hence lower than the case without air resistance. 4(b) *ar has a non constant magnitude. ar = R m For both cases, the distance travelled by the ball is the same, which means the area of the v-t graph is the same for upward and downward motion. Since ad < au, the decrease in speed on the rise is more than the increase in speed on the fall. Therefore, td is greater than tu since the two areas are the same. mg Ball moving upward under free fall Ball moving upward in the presence of air resistance v1 mg FR v2 t/s velocity / m s1 td tu Net deceleration au = g + ar Ball moving upward in the presence of air resistance R mg R v1 Net acceleration ad = g – ar Ball moving downward in the presence of air resistance R mg Rv2

Kinematics Tutorial soln 2026/JPJC/PHYSICS/9478 6 4(c) Key differences: Lower maximum height, shorter range, non-symmetrical path about the highest point. 4(d) Without air resistance KE decreases to a minimum as GPE increases to a maximum. The sum of KE and GPE is a constant. With air resistance, the total sum of KE and GPE decreases continuously. The decrease in KE is more than the increase in GPE on the rise, and the decrease in GPE is more than the increase in KE on the fall. total energy KE GPE total energy KE GPE

Kinematics Tutorial soln 2026/JPJC/PHYSICS/9478 7 5(a) (downward is positive direction) The gradient of v-t graph gives a-t graph 5(b) From 0 to P, the acceleration decreases from g because the velocity and hence the air resistance directed upward increases. The net force ( mg – air resistance) and hence the acceleration decreases, causing the velocity to increase at a slower rate, until it reaches the maximum terminal velocity at P. At this point the air resistance = mg, the acceleration is zero. From P to Q, it maintains the terminal velocity. At Q the air resistance increases greatly due to the opening of the parachute. The net force is now pointing upward, creating an upward acceleration, hence reducing the downward velocity. The velocity will continue to decrease until the air resistance is again equal to the weight at point R. From R to S, the parachutist moves with a lower terminal velocity. 5(c) Higher terminal velocities due to higher weight.

Kinematics Tutorial soln 2026/JPJC/PHYSICS/9478 8 6(a) As the object moves