How can I determine the vertical distance the ball clears the wall?

[-EquinoX-]

Homework Statement

A playground is on the flat roof of a city school, 6.00m above the street below. The vertical wall of the building is 7.00 m high forming a 1m high railing around the playground. A ball has fallen to the street below and a passerby returns it by launching it at an angle of 53 degrees above the horizontal at a point 24 m from the base of the bulding wall. The ball takes 2.20 to reach a point vertically above the wall. Find the vertical distance by which the ball clears the wall.

Homework Equations

The Attempt at a Solution

I can get the initial velocity first by plugging in the formula:

24 = Vi*cos(53)*(2.2) then

finding for Vi I get 18.1. Now how do I find the height the ball clears the wall?

 

[Kurdt]

You need an equation that has the distance the acceleration and the initial and final velocities in the y direction. You know at the max height the velocity in the y direction will be zero. You know the initial velocity and the acceleration and thus you can work out the height.

 

[baba89]

To determine the vertical distance the ball clears the wall, you can use the equation for projectile motion:

h = h0 + Vi*sin(θ)*t - (1/2)gt^2

Where h is the vertical distance, h0 is the initial height (6m in this case), Vi is the initial velocity (18.1 m/s), θ is the launch angle (53 degrees), t is the time the ball takes to reach the top of the wall (2.2 seconds), and g is the acceleration due to gravity (9.8 m/s^2).

Plugging in these values, we get:

h = 6 + (18.1*sin(53)*2.2) - (1/2)(9.8)(2.2)^2

h = 6 + 24.4 - 23.1

h = 7.3

Therefore, the vertical distance the ball clears the wall is 7.3 meters.