First slide

Question on vertical

Question

A boy sees a ball going up and then back down through a window 2.45 m high. If the total time taken by the ball is 1 sec, the height above the window that the ball rises is

Moderate

A

0 .306 m
B

3.06 m
C

0 .030 m
D

306 m

Solution

Here the ball is in sight for 1 sec. Thus the ball takes 1/2 second to go up across the window and 1/2 second to come down across it. Let h be the height that the ball rises above the window and t be the time taken to fall through h. Then the time taken to fall through the distance
$(h + 2 \cdot 45) m$ would be $(t + \frac{1}{2})$ hence
$\begin{array}{l} h = 0 \times t + \frac{1}{2} {gt}^{2} or h = \frac{1}{2} {gt}^{2} . . . . (1) \\ and (h + 2 \cdot 45) = 0 \times (t + \frac{1}{2}) + \frac{1}{2} g {(t + \frac{1}{2})}^{2} . . . . (2) \end{array}$
From eqs. (1) and (2)
$\frac{1}{2} {gt}^{2} + 2 \cdot 45 = \frac{1}{2} g {(t + \frac{1}{2})}^{2}$
Solving, this equation, we get
$\begin{array}{l} t = \frac{1}{4} \sec \\ ∴ h = \frac{1}{2} \times 9 \cdot 8 \times {(\frac{1}{4})}^{2} = 0 \cdot 306 m \end{array}$

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