A place kicker must kick a football from a point 36.0 meters (about 40yds)from the goal. as a result of the kick, the ball must clear the crossbar, which is 3.05 meters high. When kicked the ball leaves the ground with a speed of 20.0 m/s at and angle of 53 degrees to the horizontal

a) how much does the ball clear or fall short of clearing the crossbar
b) does the ball approach the crossbar while still rising or while falling?

hf=hi+ViSintheta*t-1/2 g t^2 in the vertical.

in the horizontal,
36=vi*cosTheta*t solve for t in terms of theta.
t=36/(vi*cosTheta)
Now put that into the other equation above.

hf=0+Visintheta*36/vicosTheta-1/2 g 36^2/Vi^2cos^2 theta

You know Vi, Theta, solve for hf.

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To solve this problem, we can break the initial velocity of the ball into horizontal and vertical components.

Given:
- Initial velocity (v₀) = 20.0 m/s
- Launch angle (θ) = 53 degrees

a) How much does the ball clear or fall short of clearing the crossbar?

To find out the vertical distance the ball clears the crossbar, we need to determine the maximum height it reaches and then subtract the height of the crossbar.

Step 1: Find the vertical component of the initial velocity:
Vertical component (v₀sinθ) = 20.0 m/s * sin(53°)
= 20.0 m/s * 0.8
= 16.0 m/s

Step 2: Use the kinematic equation to find the maximum height (h) reached by the ball:

v₁² = v₀² + 2aΔy

At the top of the trajectory, the vertical component of the velocity will be zero (v₁ = 0). The initial vertical velocity (v₀sinθ) would only contribute to the decrease in height.

0 = (16.0 m/s)² + 2(-9.8 m/s²)Δy

Solving for Δy (vertical displacement):

Δy = (16.0 m/s)² / (2 * 9.8 m/s²)
≈ 13.0 m

Step 3: Calculate the difference between the maximum height (13.0 m) and the height of the crossbar (3.05 m):

Difference = 13.0 m - 3.05 m
≈ 9.95 m

Therefore, the ball clears the crossbar by approximately 9.95 meters.

b) Does the ball approach the crossbar while still rising or while falling?

The ball will approach the crossbar while falling since its vertical velocity decreases until it becomes zero at the top of the trajectory and then increases in the opposite direction due to gravity.