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

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edwinandrew
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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=[itex]\Delta[/itex]x/[itex]\Delta[/itex]

this is for vertical motion

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

Vfy^2=Viy^2+2ay[itex]\Delta[/itex]y
[itex]\Delta[/itex]y=Vfy[itex]\Delta[/itex]t-1/2ay([itex]\Delta[/itex]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.
 
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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.
 
jackarms said:
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
 
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.