You throw a vall toward a wall with a speed of 25m/s and at an angleof 40 degree above the horizontal. The wall is 22m from the realease point of the ball. a) How far above the release point does the ball hit the wall? b) What are the horizontal and vertical components of its velocity as it hits the wall? c) When it hits has it passes the highest point on its trajectory?

Question

You throw a vall toward a wall with a speed of 25m/s and at an angleof 40 degree above the horizontal. The wall is 22m from the realease point of the ball. a) How far above the release point does the ball hit the wall? b) What are the horizontal and vertical components of its velocity as it hits the wall? c) When it hits has it passes the highest point on its trajectory?

I am confused...

Answer 1

No worries, I'm here to help! Let's break it down step by step.

a) To find out how far above the release point the ball hits the wall, we need to calculate the vertical distance traveled by the ball.

First, we can find the time it takes for the ball to reach the wall using the horizontal component of its initial velocity. The horizontal velocity (Vx) can be found by multiplying the initial velocity (25 m/s) by the cosine of the launch angle (40 degrees).

Vx = 25 m/s * cos(40°)

Once we have the horizontal velocity, we can calculate the time (t) it takes for the ball to reach the wall using the formula:

distance = velocity * time

Rearranging the equation to solve for time:

time = distance / velocity

Since the distance is given as 22 meters and the velocity in the horizontal direction is Vx, we have:

t = 22 m / Vx

Now, let's calculate the vertical displacement (Δy) using the formula:

Δy = V_initial * sin(θ) * t + (1/2) * g * t^2

where V_initial is the initial vertical velocity (which can be found by multiplying the initial velocity of 25 m/s by the sine of the launch angle), θ is the launch angle (40 degrees), t is the time it takes to reach the wall, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Δy = (25 m/s) * sin(40°) * t + (1/2) * (9.8 m/s^2) * t^2

Now, you can substitute the value of t that we calculated earlier and solve for Δy.

b) To find the horizontal and vertical components of the ball's velocity when it hits the wall, we can use the equations:

Vx = initial velocity * cos(θ)
Vy = initial velocity * sin(θ)

Here, Vx is the horizontal component of the velocity and Vy is the vertical component.

c) When the ball hits the highest point on its trajectory, its vertical velocity (Vy) becomes zero. You can use this fact to determine the time it takes for the ball to reach its highest point. Consider the equation:

Vy = initial velocity * sin(θ) - g * t

Set Vy to zero and solve for t to find when the ball reaches its highest point.