Question

ANALYSIS

A rock is thrown vertically upward with with a speed of 12m/s. Exactly 1s later a ball thrown up vertically along the same path with a speed 18m/s.

1). At What time will the collision occur?

2). At what height will the collision occur?

3). Answer (1) and (2), assuming if the order is reversed: the rock is thrown 1.00s before the ball.

Answer 1

1) 2)

$$12t-(0.5)9.8t^2=18(t-1)-(0.5)9.8(t-1)^2\\t=1.45\ s\\h=12(1.45)-(0.5)9.8(1.45)^2=7.1\ m$$

3)

$$12(t-1)-(0.5)9.8(t-1)^2=18t-(0.5)9.8(t)^2\\t=4.45\ s\\h=18(4.45)-(0.5)9.8(4.45)^2=17\ m$$