Answer :
Start by finding the time interval, delta t, that it takes for the item to hit the wall. You can do so by knowing that t = d/s. We can solve for s, which is speed, by finding the initial velocity in the horizontal direction.
We can find this by using the launch angle given, 30 degrees, and solving for its horizontal component (parallel to x-axis).
This is found by cos30 = v_x/20
Solve for v_x to get: 17.32051 m/s. Now we can use this initial velocity in the horizontal direction in the equations.
Time = distance/speed
Time = 25 m/17.32051 m/s = 1.44338 seconds
Now we have t = 1.44338 seconds
We want to find the height of the item at a certain time, this time is 1.44338 seconds. We don't have the final velocity, so therefore, we can use the kinematic equation for constant acceleration (object in projectile motion has constant acceleration of -9.8 m/s^2 due to gravity):
No Final Velocity
x_f = x_i + v_i * t + 1/2a (t^2)
x_f and x_i are final and initial positions, respectively, and v_i is the initial velocity. The initial position of this item is 0 m, since we are starting from ground level.
We are trying to solve for the final position, x_f, and we know the variables v_i (17.32051 m/s), t (1.44338 s), and a (9.8 m/s).
Substitute these values into the equation to find the final position:
x_f = (0) + (17.32051 m/s)(1.44338 s) + 1/2(9.8 m/s^2)(1.44338 s)^2
x_f = (25.00007772) + (10.20839454)
x_f = 35.20847227
Now we know that at the time it takes for the item to hit the wall, 1.44338 s, the item's height is 35.20847 meters.