Question

a football player kicks a ball at an angle of 30degree with the horizontal with the velocity of 60m/s .The maximum height reached by the ball is​

Answer 1

Explanation:

Answer :

The maximum hieght reached by the ball is 45 m.

Explanation :

Given :

Angle of projection, θ = 30°

Intial velocity of the ball, u = 60 m/s

Final velocity of the ball, v = 0 m/s

[Final velocity of a body at maximum height is 0, v = 0]

Acceleration due to gravity, g = 10 m/s² (Approx.)

[Acceleration due to gravity = 9.8 m/s²]

To find :

Maximum height reached by the ball, h = ?

Knowledge required :

Formula for maximum height in case of a projectile :

$\boxed{\sf{h_{(max.)} = \dfrac{u^{2}sin^{2}\theta}{2g}}}$

Where,

h = Maximum height

u = Initial velocity

θ = Angle of Projection

g = Acceleration due to gravity

Solution :

By using the formula for maximum height in case of a projectile and substituting the values in it, we get :

$:\implies \sf{h_{(max.)} = \dfrac{u^{2}sin^{2}\theta}{2g}}$

$:\implies \sf{h_{(max.)} = \dfrac{60^{2} \times sin^{2}30^{\circ}}{2 \times 10}}$

$:\implies \sf{h_{(max.)} = \dfrac{60^{2} \times \bigg(\dfrac{1}{2}\bigg)^{2}}{20}}$

$:\implies \sf{h_{(max.)} = \dfrac{60^{2} \times \dfrac{1}{4}}{20}}$

$:\implies \sf{h_{(max.)} = \dfrac{3600 \times \dfrac{1}{4}}{20}}$

$:\implies \sf{h_{(max.)} = \dfrac{3600}{20 \times 4}}$

$:\implies \sf{h_{(max.)} = \dfrac{3600}{80}}$

$:\implies \sf{h_{(max.)} = \dfrac{360}{8}}$

$:\implies \sf{h_{(max.)} = 45}$

$\boxed{\therefore \sf{h_{(max.)} = 45\:m}}$

Therefore,

Maximum height reached by the ball, h = 45 m.

Answer 2

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