Question

A 150 g cricket ball is hit by a batsman with an initial
velocity of 40 m/s at an angle 60° with the horizontal. Find
its kinetic energy when it reaches its maximum height.

Answer 1

Answer:

30J

Explanation:

At maximum height there is only horizontal component of velocity because there is no acceleration in x direction if we take this motion in x-y plane ,so as the ball has only horizontal velocity which is 20m/s then the K.E of the ball is=1/2×0.15×(20)²=30J

Answer 2

The kinetic energy of the ball when it reaches its maximum height is 30 J.

Explanation:

Given the mass of the ball is , m = 150 g = 0.15 kg.

The horizontal component of velocity , $v_x=vcos\theta$.

Here given $v=40m/s, \theta=60^0$.

so $v_x=40(cos60)=20m/s$.

Now the vertical component of velocity will be 0 at max height.

The kinetic energy is given as

$k=\frac{1}{2}mv^2$

Only horizontal component of velocity is there at the max height.

substitute the values, we get

$k=0.5\times0.15kg\times(20m/s)^2$

k = 30 J.

Thus, the kinetic energy of the ball when it reaches its maximum height is 30 J.

#Learn More: Kinetic energy.

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