Questions for Kinematics (JC) Set 1

A rescue plane is flying horizontally with a speed of 50 m s-1 and at an altitude of 200 m above the sea when it drops a warning flare.

Neglecting air resistance and assuming that the plane does not change its course, speed or altitude, how far from the plane is the flare when it hits the water?

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The flare will be released with the same horizontal velocity of 50 m s-1. Since there is no acceleration in the horizontal direction, the flare will be directly below the position of the plane when it hits the water

Answer: 200 m


A boy throws a ball vertically upwards. The ball falls back into his hand 2.0 seconds later. Neglecting air resistance, what is the speed at which the ball leaves his hand?

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Taking upward motion as positive,
s = ut + ½ at2
0 = u(2) + ½ (-9.81)(22)
u = ½ (9.81)(2)
u = 9.81 m s-1


A ball was kicked over a 8.0m wall as shown below with a velocity u at an angle of 50° above the horizontal. At the highest point of the trajectory, the ball managed to just go over the wall. It landed into a depression 2.0 m deep.

ball over wall qn

a) Express the initial horizontal velocity and initial vertical velocity in terms of u.

b) Calculate the initial speed of the ball
c) Calculate the vertical velocity of the ball vy just before it hits the depression.

d) Hence, or otherwise, calculate the velocity of the ball v just before it hits the depression, and the angle it makes below the horizontal.

e) Calculate the time of flight

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a) ux = u cos 50°
uy = u sin 50°b) Taking upwards and rightwards as positive, consider the ball at its highest point of its trajectoryvy2 – uy2 + 2aysy
0 = (u sin 50°)2 + 2(-9.81)(8.0)
u = 16.355
u = 16 m s-1
vy2 – uy2 + 2aysy
vy2 = (16.355 x sin 50°)2 + 2(-9.81)(-2.0)
vy2 = 196.2
vy = -14.007
vy = -14 m s-1

d) v2 = vx2 + vy2
v2 = (16.355 cos 50°) + (-14.007)2
v = 17.513
v = 17 m s-1

tan θ = (vy / vx)
tan θ = (14.007 / 16.355 cos 50°)
θ = 53.111°
θ = 53°

The velocity of the ball is 17 m s-1 at an angle of 53° below the horizontal.

e) vy = uy + ayt
-14.007 = (16.335 x sin 50°) + (-9.81)
t = 2.7049
t = 2.7 s


The acceleration-time graph of an object moving in a straight line is as shown. The object started its motion from rest.

acceleration-time graph qn

At which point is the body moving with the largest speed?

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Area under acceleration-time graph = change in velocity of the object.At point 2, the area under the graph is the largest.

Answer: 2

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