A player throws a ball at an angle of 30 up with the horizontal with a velocity of 14m/s . If the point of projection is at a height of 12m from the ground, calculate the distance up to which the ball is thrown by the player?

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A player throws a ball at an angle of 30 up with the horizontal with a velocity of 14m/s . If the point of projection is at a height of 12m from the ground, calculate the distance up to which the ball is thrown by the player?

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Problem: A player throws a ball at an angle of 30 up with the horizontal with a velocity of 14m/s . If the point of projection is at a height of 12m from the ground, calculate the distance up to which the ball is thrown by the player?

Solution:

When an object is thrown at an angle to the horizontal, it follows a parabolic path. The two independent motions are horizontal and vertical. The horizontal motion is uniform and the vertical motion is accelerated motion due to gravity. We can resolve the initial velocity into horizontal and vertical components.

Given:

Angle of projection, θ = 30°
Initial velocity, u = 14m/s
Height of projection, h = 12m
Acceleration due to gravity, g = 9.8m/s²

Resolving the initial velocity into horizontal and vertical components

The horizontal component of the initial velocity is given by:

u_x = u cos θ

u_x = 14 cos 30°

u_x = 12.124 m/s

The vertical component of the initial velocity is given by:

u_y = u sin θ

u_y = 14 sin 30°

u_y = 7 m/s

Calculating the time taken to reach maximum height

The time taken to reach the maximum height is given by:

t_max = u_y / g

t_max = 7 / 9.8

t_max = 0.71 s

Calculating the maximum height reached by the ball

The maximum height reached by the ball is given by:

h_max = u_y² / 2g

h_max = 7² / 2 x 9.8

h_max = 2.537 m

Calculating the horizontal distance traveled by the ball

The horizontal distance traveled by the ball is given by:

R = u_x x t_max

R = 12.124 x 0.71

R = 8.61 m

Calculating the total distance traveled by the ball

The total distance traveled by the ball is given by:

D = R + x

where, x is the distance traveled by the ball while going up and coming down.

x = 2h_max

x = 2 x 2.537

x = 5.07 m

D = R + x

D = 8.61 + 5.07

D = 13.68 m

Therefore, the ball is thrown up to a distance of 13.68 m by the player.

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