Theory question 115
Dynamics(a) A ball is thrown from the ground upwards with a velocity of 30ms−1. A second later, another ball held at rest is released from a height of 50m to hit the first ball. Find the: (i) time when the balls meet
(ii) height at which the balls meet [Take g = 10ms−2]
(b) A bullet was released from a hunter's gun with an initial velocity of 8ms−1. It then accelerates at 3.5ms−2 for 15 seconds. Find the (i) final velocity of the bullet
(ii) distance travelled by the bullet
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(a) Height of first ball after t s: h = 30t − 5t². Second ball (released at t = 1): h = 50 − 5(t − 1)².
(i) 30t − 5t² = 50 − 5t² + 10t − 5, so 20t = 45 and t = 2.25 s after the first ball is thrown (1.25 s after the second is released).
(ii) h = 30(2.25) − 5(2.25)² = 67.5 − 25.3 ≈ 42.2 m.
(b)(i) v = u + at = 8 + 3.5 × 15 = 60.5 ms⁻¹.
(ii) s = ut + ½at² = 120 + ½ × 3.5 × 225 = 513.75 m.
Measure both balls' heights from the ground with the same clock.