Question

A ball is projected horizontally from the top of a cliff
on the surface of the Earth with a speed of 15 ms-1.
Assuming, that there is no air resistance, what will its
speed 2 s later? Take acceleration due to gravity as
10 ms-2​

Answer 1

Answer:

Hortizontal velocity will not alter and thus will be 40 m/s. Now considering the vertical velocity, after 3 seconds because of accelearation due to gravity the vertical velocity would be ‘gt' that is 10 (approx) m/s^2 × 3 s= 30 m/s. Now we have both vertical and horizontal components. 40 as horizontal and 30 as vertical. The resultant would be 50 m/s with an angle of 37 degree to horizontal.

Explanation:

Answer 2

Speed of the ball is 25m/s.

Given :

ux (horizontal velocity) = 15 m/s

Time (T) = 2s

acceleration due to gravity (g) = 10m/s^2

To find :

Its speed after 2s

Solution :

Let x be the horizontal direction

ax = 0

Vx = ux = 15m/s

y be the vertical direction

uy = 0

ay = g = 10m/s^2

Using formula,

Vy = uy + ayT

Vy = 0 + 10 × 2 = 20m/s

Speed of the ball V will be equal to ,

= √(Vx ^2 + Vy^2)

= √(( 15)^2 + (20)^2)

= √(225 + 400)

= √625

= 25

Hence, speed of the ball is 25m/s.

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