A ball is dropped from the top of building .The ball takes 0.5 sec to fall past the 3 m length of a window some distance from the top of building ,How fast was the ball going as it passes the top of the window ? How far is top of the window from the point at which ball was dropped ? ( take g = 9.8 m/s² )

Open in App

Solution

Let, ‘v_t’ be the speed of the ball at the top point of the window, ‘v_b’ be the speed at the bottom point of the window. The acceleration we know is ‘g’, g = 9.8 m/s²
The time taken to cross the window, t =0.5 s and the length of the window is S = 3 m.
Using,
S = ut + ½ at²
=> 3 = v_t × t + ½ × g × t²
=> 3 = v_t × 0.5 + ½ × 9.8 × 0.5²
=> v_t = 3.55 m/s
When the ball was just dropped its initial velocity was 0. So, the distance after which the velocity is v_t = 3.55 m/s is (it is the distance from the point from which the ball is dropped to the top of the window),
v² = u² + 2as
=> 3.55² = 0 + 2 × 9.8 × s
=> s = 0.64 m

Suggest Corrections

Similar questions

Q. A ball is dropped from the top of a building. The ball takes

0.5

s to fall past the

3

m length of a window at certain distance from the top of the building.
(

g = 10 m / s^{2}

)
How far is the top of the window from the point at which the ball was dropped ?

Q. A ball is dropped from the top of a building. The ball takes

0.5 s

to fall past the window of

3 m

in length at certain distance from the top of the building. (

g = 10 m / s^{2}

) How fast was the ball going as it passed the bottom of the window?

Q. A ball is dropped from the top of a building. The ball takes

0.5

s to fall past the length of a window, the top of the window being at a distance of

3

m from the top of the building. If the speed of the ball at the top and the bottom of the window are

V_{T}

and

V_{B}

respectively, then

(g = 9.8 m / s e c^{2})

Q. A ball is dropped from the top of a building. The ball takes

0.5

s to fall past the window

3

m in length at certain distance from the top of the building. (

g = 10 m s^{- 2}

)
Speed of the ball as it crosses the top edge of the window is

A ball is dropped from the top of a building. The ball takes 0.5sec to fall past the 3m height of a window some distance from the top of the building. If the speed of the ball at the top and the bottom is $V_{T}$ and $V_{B}$ , the establish a relationship between vb and vt take (g = 9.8m/s).

Join BYJU'S Learning Program