A player throws a ball towards a high wall. The ball leaves the player's hand at a height of 2 meters above the ground with a velocity of 5.00 m/s at an angle of 30°. The wall is at a horizontal distance 4.00 meters away from the player. At what height above the ground will the ball hit the wall?

A player throws a ball towards a high wall. The ball leaves the player's hand at a height of 2 meters above the ground with a velocity of 5.00 m/s at an angle of 30°. The wall is at a horizontal distance 4.00 meters away from the player. At what height above the ground will the ball hit the wall?

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Question

A player throws a ball towards a high wall. The ball leaves the player's hand at a
height of 2 meters above the ground with a velocity of 5.00 m/s at an angle of 30°. The
wall is at a horizontal distance 4.00 meters away from the player. At what height
above the ground will the ball hit the wall?
Wall
Y=2 m
X=4 m

Expert Solution

Step 1

Projectile

A projectile is an object that is thrown without any fuel. Generally, a projectile is thrown with a velocity that makes an angle with respect to the horizontal. Examples of a projectile are a meteor falling from the sky and an artillery gun firing towards enemy positions.

Let us consider that a projectile is thrown with a velocity $v$ at an angle $θ$ with respect to the horizontal above a height $h$

The component of velocity along the X and Y directions are

$v_{x} = v \cos θ v_{y} = v \sin θ$

The X coordinate of the projectile at any later time $t$ considering the base of the building as the origin

$x = v_{x} t = v \cos θ t$

The Y coordinate of the projectile at any later time $t$ is

$y = h + v_{y} t - \frac{1}{2} g t^{2} \Rightarrow y = h + v \sin θ t - \frac{1}{2} g t^{2}$

From the parametric equation for X direction

$t = \frac{x}{v \cos θ}$

Substituting this in the Y equation

$y = h + v \sin θ \times (\frac{x}{v \cos θ}) - \frac{g}{2} \times {(\frac{x}{v \cos θ})}^{2} \Rightarrow y = h + x \tan θ - \frac{g x^{2}}{2 v^{2} \cos^{2} θ}$

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