Two Dimensional Kinematics

Question

A ball is kicked from the ground at an angle of θ = 57 degrees with respect to the horizontal. The ball is in the air for a time t_m = 2.1 s before it lands back on the ground.

Numerically, what is the total horizontal distance, d_m in meters, traveled by the ball in the time, t_m?

I've gotten 11.78 and it's being marked wrong, either rounded incorrectly somewhere or it's completely off.

Answer 1

Another way to it is to take note of the fact that at its highest point the balls vertical velocity is zero and the time equal to one half of the time when it hits the ground.

The equation for the vertical is 0 = v sin 57 - gt/2, where v is the initial velocity and t = 2.1

Solve this to get v and then use the horizontal equation of motion

d = vcos57 x (2.1)

I got d = 45.3, using g = 10 m/s/s

Answer 2

Hello Justin, did you answer your own question? If not, here are some pointers. I urge you to look at your basic kinematic constant acceleration motion equation:

y_f = y₀ + v₀ t + 0.5 a t²

If the ground is y = 0, then what do we find as the initial vertical velocity component for the ball?

After we find that, we use trigonometry to determine the horizontal velocity component (which is constant through the trajectory).

Multiplying by the time in the air, you should find your horizontal travel distance.

If you need more clarification or if you have further questions, please feel free to message me for a lesson!