How Do You Calculate the Initial Velocity and Range of a Soccer Ball?

[edwinandrew]

Homework Statement

A person kicks a soccer ball with an intial velocity directed 53 degrees above the horizontal. The ball lands on a roof 7.2 m high. The wall of the building is 25m away, and it takes the ball 2.1 seconds to pass directly over the wall. Calculate the intial velocity of the ball an its horizontal range.

Homework Equations

this is for horizontal motion

Vix=$\Delta$x/$\Delta$

this is for vertical motion

Vfy=Viy+ay($\Delta$t)
$\Delta$y=viy$\Delta$t+1/2ay($\Delta$t)^2

Vfy^2=Viy^2+2ay$\Delta$y
$\Delta$y=Vfy$\Delta$t-1/2ay($\Delta$t)^2

The Attempt at a Solution

Okay hi everyone this is my first post! Well okay i don't understand the part about how it takes 2.1 seconds to go over the wall because in most of the questions i have done they give total time.

 

[jackarms]

Welcome to Physics Forums -- from the problem, I assume that means it takes 2.1 seconds to get to the wall, and then it travels for a bit to get all the way to the wall. You could handle that in a couple different ways -- say finding the height at that point, the distance it travels from that point along the wall, and the time that the second leg of the motion would take, etc.

 

[edwinandrew]

Welcome to Physics Forums -- from the problem, I assume that means it takes 2.1 seconds to get to the wall, and then it travels for a bit to get all the way to the wall. You could handle that in a couple different ways -- say finding the height at that point, the distance it travels from that point along the wall, and the time that the second leg of the motion would take, etc.

Thats sounds really confusing

 

[jackarms]

Just handle the problem in two parts. One part is the ball launched to that point above the wall, and the other part is the ball falling from that point down onto the roof. See where the analysis of that gets you.

 

[HalfLostAndSearching]

Hello! I would approach this problem by breaking it down into two separate components: the horizontal motion and the vertical motion of the soccer ball.

For the horizontal motion, we can use the equation Vix = Δx/Δt to solve for the initial velocity (Vix) of the ball. We know the distance (Δx) is equal to 25m and the time (Δt) is 2.1 seconds. Therefore, Vix = 25m/2.1s = 11.9 m/s.

For the vertical motion, we can use the equations Vfy = Viy + ay(Δt) and Δy = ViyΔt + 1/2ay(Δt)^2 to solve for the initial vertical velocity (Viy) and the maximum height (Δy) of the ball. We know that at the maximum height, the vertical velocity is 0, so we can use the equation Vfy^2 = Viy^2 + 2ayΔy to solve for Viy. Plugging in the given values, we get 0 = Viy^2 - 2ay(7.2m). Since we know that the initial angle is 53 degrees above the horizontal, we can use trigonometry to solve for Viy, which is equal to 11.6 m/s. Plugging this value into the equation Δy = ViyΔt + 1/2ay(Δt)^2, we can solve for the maximum height, which is 4.9m.

Therefore, the initial velocity of the ball is 11.9 m/s and its horizontal range is 25m. I hope this helps! Let me know if you have any other questions.