Question

20.
A ball is projected from the top of a building of height 45 m, with a speed of 40 ms', parallel to the
ground. Find the speed with which it hits the ground.
1) 80 ms
2) 60 ms1
3) 70 ms 1
4) 50 ms -1​

Answer 1

Answer:

v²=u²+2as

v²=40²+2×10×45

  =1600+900

v²=2500

∴v=50 m/s

Answer 2

Answer:

The speed at which the ball hits the ground is 50m/s.

Explanation:

Given that,

A ball is projected from the top of a building of a height of 45m, with a speed of 40ms, parallel to the ground and we are required to find the speed with which it hits the ground.

The time of flight of the ball is $t=\sqrt{\frac{2h}{g}}$

We know that $h=45m,g=10m/s^2$

Therefore,

$t=\sqrt{\frac{2\times45}{10}}=3\:sec$

Using the equation of motion for x-direction we get

$v_{x}=u_{x}+a_{x}t=40+0=40m/s$

Using the equation of motion for x-direction we get

$v_{y}=u_{y}+a_{y}t\\=0+(-10)(3)=-30m/s$

Hence, The resultant speed is

$v=\sqrt{v_{x}^2+v_{y}^2}=\sqrt{40^2+(-30)^2}\\=\sqrt{2500}=50m/s$

Therefore,

The speed at which the ball hits the ground is 50m/s.

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