Q#16 (Past Exam Paper – June 2016 Paper 11 Q7)
A student uses a spring gun to launch a steel ball with a constant horizontal velocity. He varies the height h of the gun and measures the horizontal displacement r of the ball when it hits the ground.
Which graph shows the variation with height h of the horizontal displacement r ?
Solution:
Answer: A.
In this question, we need to consider the horizontal and vertical motion of the steel ball individually (projectile motion).
The horizontal motion is NOT affected by the acceleration due to gravity. So, the horizontal speed of the ball is constant.
Speed = Distance / time
vh = r / t
Horizontal displacement, r = vh t ----------- (1)
The horizontal and vertical motion can be considered separately. But, these can be related to each other by the time t. As the ball falls a vertical distance h in a time t, it also moves the horizontal distance r.
For the vertical motion,
s = ut + ½ at2
Since the ball is initially at rest, u = 0. Also, here s = h and acceleration a = g.
h = ½ gt2
Time t = √(2h/a) ----------- (2)
Replacing eqn (2) in eqn (1), we can have an equation that relates r and h.
r = vh ×√(2h/a)
So, r ∝ √h
The horizontal displacement r is proportional to the square root of the vertical displacement h.
(From Maths, if you plot the graph of y = √x,) This corresponds to shape in graph A.
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