You are 75m up high and are tossing footballs to friends. If you throw it horizontally with a speed of 26m/s, how close to the building (horizontally) must your

Question

This is everything that is provided in the question aside from a few word changes due to the academic honesty policy and the character limit. To clarify the problem, you are 75m high because you are on top of a balcony.

I tried to find the x and y components to solve the problem through trigonometry, but my answer seems incorrect. I also used kinematic equations to solve for time but then I get stuck because I do not know if the 26m/s speed applies to both horizontal and vertical speed. Lastly, this question does not take air resistance into account so hopefully, this simplifies the problem.

Answer 1

There are two motions here, vertical and horizontal.

 

Both begin and end at the same time

 

Vertical:

 

h - 75 m

g = -9.8 m/s

t = ?

 

h = 1/2 * g * t^2 (the initial velocity in the vertical direction is zero)

 

t = (2h/g)^1/2 = (150/9.8)^1/2 = 3.9 seconds

 

Horizontal: (there is no acceleration in the horizontal direction)

 

displacement = velocity*time = 26 m/s * 3.9 s = 101.7 m (the ball lands 101.7 m from the base of the balcony)