First slide

The 3 equation for uniform acc

Question

A person sitting in the ground floor of a building notices through the window of height 1.5 m, a ball dropped from the roof of the building crosses the window in 0.1 s. What is the velocity of the ball when it is at the topmost point of the window? (Take, g = 10 m/s²) [NEET 2020]

Moderate

A

15.5 m/s
B

14.5 m/s
C

4.5 m/s
D

20 m/s

Solution

According to question, time taken by the ball to cross the window,
t = 0.1s, h = 1.5m
If u be the velocity at the top most point of the window, then from equation of motion,
$h = u t + \frac{1}{2} g t^{2}$
$\Rightarrow 1.5 = u \times 0.1 + \frac{1}{2} \times 10 \times (0.1)^{2}$
$\Rightarrow 1.5 = 0.1 u + 0.05$
$\Rightarrow u = \frac{1.5 - 0.05}{0.1} = \frac{1.45}{0.1} = 14.5 m / s$

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