A boy throws a ball into air at 45^ from the horizontal to land it on a roof of a building of height H . If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of H is

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 15

For projectile motion at 45°, vertical component of initial velocity:

v_{0y} = v_0 \sin 45° = \frac{v_0}{\sqrt{2}}

Maximum height occurs at

t = 2

s when

v_y = 0

:

0 = v_{0y} - gt

\Rightarrow v_{0y} = 10 \times 2 = 20

m/s
Thus

\frac{v_0}{\sqrt{2}} = 20

, so

v_0 = 20\sqrt{2}

m/s
Height at landing time

t = 3

s:

H = v_{0y}t - \frac{1}{2}gt^2 = 20 \times 3 - \frac{1}{2} \times 10 \times 9

= 60 - 45 = 15

m

A boy throws a ball into air at 45^ from the horizontal to land it...

Options

Correct answer

Step-by-step solution

More from Motion in Two Dimensions