Q. 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 ____ m.
\((g = 10 \text{ m/s}^2)\)
Explanation
Time to reach maximum height is given as 2 s.
For vertical motion,
\[
t = \frac{u_y}{g}
\]
So,
\[
u_y = g t = 10 \times 2 = 20 \text{ m/s}
\]
Angle of projection is 45°, hence
\[
u_x = u_y = 20 \text{ m/s}
\]
Vertical displacement after time \(t = 3\) s is given by
\[
y = u_y t - \frac{1}{2} g t^2
\]
Substitute values:
\[
y = 20 \times 3 - \frac{1}{2} \times 10 \times 3^2
\]
\[
y = 60 - 45 = 15
\]
Thus, the height of the building is
\[
H = \boxed{15 \text{ m}}
\]
Hence, the correct answer is 15.