Solving a Soccer Player's Physics Problem: Horizontal Kick Speed

AI Thread Summary
A soccer player kicks a rock horizontally from a 40m cliff, and the sound of the impact is heard 3 seconds later, leading to a calculation of the initial speed of the rock. With the speed of sound at 343 m/s, the player’s initial kick speed is determined to be approximately 9.3 m/s. In a separate scenario, a basketball player jumps 2.80m horizontally and reaches a maximum height of 1.85m, prompting calculations for time of flight and velocity components. The player’s vertical velocity at takeoff is calculated to be around 4.033 m/s, with a takeoff angle of approximately 50.87 degrees. Both problems involve applying kinematic equations to solve for unknowns in projectile motion.
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Homework Statement


A soccer player kicks a rock horizontally off a cliff 40m high into a pool of water. If the player hears the sound 3 seconds later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

Homework Equations


g=9.8


The Attempt at a Solution


I think I got it, but can you guys check? Is the answer 9.3 m/s?



Homework Statement


A basketball star covers 2.80m horizontally in a jump to dunk the ball. His motion through space can be modeled as that of a particle at his center of mass. His center of mass is at elevation 1.02m when he leaves the floor. It reaches a maximum height of 1.85m above the floor and is at an elevation of 0.900m when he touches down again. Determine his time of flight and his horizontal and vertical velocity components at the instant of takeoff and his takeoff angle.


Homework Equations


Kinematic Equations


The Attempt at a Solution


I think I might have misunderstood the problem, but here I go.
First I translated everything down -.9 to make it easier on myself.
vxt=2.8
y=vyt-4.9t^2+.12

Ok I set the vY=0 so 0=vy-9.8t. t=vy/9.8, I put this back into the y position equation and set that equal to .95. I got vy=4.033. Then I get y=4.033t-4.9t^2+.12 and I set that equal to 0. Therefore t=.8518 Solving for vx(.8518)=2.8, I get vx=3.28. Then if I was right about this, the takeoff angle is 50.87. Am I right?
 
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For 1, I get roughly that number, but maybe I carried more precision.

For 2, I didn't run the quadratic, but the equation looks ok.
 
Ok thanks man!
 
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