Question

Two tall buildings are 30 m apart. The speed with which a ball must be thrown horizontally from a window 150 m above the ground in one building so that it enters a window 27.5 m from the ground in the other building is

Answer 1

Answer:

6 m/s

Time to reach ball to 27.5 m from ground or

150 - 27.5 = 122.5 m from roof . Note that vertical velocity is initially zero

$s = ut + \frac{1}{2} a {t}^{2} \\ \\122.5 = 0(t) + \frac{1}{2} \times 9.8 \times {t}^{2} \\ \\ {t}^{2} = \frac{122.5}{4.9} = 25 \\ \\ t = \sqrt{25} = 5 \: s$

Now in this time in horizontal direction ball must have travelled 30 m to enter the window , so

$u = \frac{30}{5} = 6 \: m {s}^{ - 1}$

Answer 2

Two tall buildings are 30 m apart.

Assume that one building is A and another is B. Distance between A and B building is 30 m.

The speed with which a ball must be thrown horizontally from a window 150 m above the ground in one building so that it enters a window 27.5 m.

From building A a ball is thrown from a window which is 27.5 m above the ground to the building B.

The ball enters the window of the building B which is 150 m above.

Total height covered = 150 - 27.5 = 122.5 m

Now,

R = u√(2h/g)

R = horizontal distance = 30 m, h = total height covered = 122.5 m and g = acceleration due to gravity = 9.8 m/s²

→ 30 = u√(2*122.5/9.8)

→ 30 = u√25

→ 30 = 5u

→ 30/5 = u

→ 6 = u

Therefore, the velocity of the ball is 6 m/s.