How far above ground level does the ball hit the fence?

Question

A ball leaves the bat with a speed of 40 m/s and an angle of 37degrees above the horizontal. A very high fence is located at a horizontal distance of 128m from the point where the ball is struck. Assuming the ball leaves the bat near ground level,.....

I got 17.07 m and just using this to check my answer.

Answer 1

First divide up the initial velocity into its horizontal and vertical components.
V is the initial velocity of 40 m/s.
Theta is the launch angle of 37 degrees above the horizon.
V_x_i and V_y_i are the horizontal and vertical components of the initial velocity respectively.

V_x_i = V * cos(theta) = 40 * cos (37) = 31.95 m/s
V_y_i = V * sin(theta) = 40 * sin (37) = 24.07 m/s

Although the vertical component of the object's velocity will change as the object experiences gravitational acceleration toward the ground, neglecting air resistance, the horizontal component of velocity can be treated as a constant.

V_x = V_x_i = 31.95 m/s
V_y = V_y_i - g*t
where t is the time after the object is launched and g is the gravitational acceleration experienced by the object (g = 9.81 m/s^2).

If the fence which the object hits is 128 meters away (in the +X direction), then the object (traveling at 31.95 m/s in the +X direction) will hit it at about 4.01 seconds after launch.

By taking this value (4.01 s) and plugging it in for t in the below shown equation to solve for the distance relative to the initial launch height the object has traveled,
d = V_y_i * t + 1/2 (g)t^2
one can calculate that the object is at a height of about 17.65 meters above its original launch height.

So I calculate the impact height to be 17.65 meters.

Answer 2

I got 17.6
Used 9.8 for g, Vyo = 24, Vxo = 32

t = 128/32 = 4 sec

y = 24t - (gt^2)/2 = 17.6