Question

A ball A is thrown straight up from the edge of the roof of a building. Another ball B is dropped from the roof 1.00 s later. You may ignore air resistance, (a) If the height of the building is 20.0 m, what must the initial speed of ball A be if both are to hit the ground at the same time? (b) On the same graph, sketch the position and velocity of each ball as a function of time, measured from when the first ball is thrown and taken origin at the ground.

Answer 1

Let ball A is thrown upward at t = 0. It takes time T to reach the ground. Ball B is dropped from the roof 1.00 s later. If both the balls reach the ground simultaneously, the flying time for ball B will be (T- 1) s.
The displacement of both the balls is (= -20 m).
Using y = $y_0$ + ut + (1/2)a$t^2$ initially for both balls are at $y_0$ = 20 and when both reaches ground where y = 0.
For ball A: 0 = 20 + $v_{0}T$ + (1/2)(-10)$T^2$ ...(i)
For ball B: 0 = 20 + (1/2)(-10)$(T-1{)}^2$ ...(ii)
From Eqs. (i) and (ii),
$v_{0}T$ = 5(2T - 1) ...(iii)
From Eqs. (i) and (iii), we get
$T^2$ - 2T - 3 = 0 $\Rightarrow$ (T + 1)(T - 3) = 0
T = 3s
From Eqs.(iii) $v_0$ = 25/3 m/s
Let ball A reaches maximum height at time t' after throwing. For ball A at maximum height, the velocity will be zero.
Using v = u + at
0 =(25/3) - 10 $\times t^{\prime }\Rightarrow t^{\prime }=$5/6 s
For maximum height reached by ball A,
Using $v^2=u^2+2as$
0 = ${\left(\frac{25}{3}\right)}^2$ - 2 x 10 x h'
h' = 125/36 m
Hence, maximum height reach by ball A
H = 20 + (125/36) = 845/36 m