Physics Footbal Punt

Question

A football is punted at 21.0 m/s at an angle of 30.0∘ above the horizon. Assume it starts 1.00 m above ground level. (Neglect any effects due to air resistance.)

What is the magnitude of vx and the magnitude of vy of the ball when it is 5.00 m above ground level? (Express your answer to three significant figures.)

Answer 1

Hi Lyric A.,

We can find the velocity in the x direction by multiplying the initial velocity by the cosine of the angle;

v_x = v_o*cos(θ) = 21.0 m/s *cos(30.0^o) = 18.2 m/s. This velocity is constant in the x direction, no matter the height.

To find the velocity in the y direction we can use the equation, v_y = v_oy +at, where v_oy = v_o*sin(θ) and a = -g. We need to know the time (t) and we can use the equation y = y_o + v_oy*t + 1/2 at², with y being positive up (a = -g = -9.81), the initial height y_o = 1.00 m, the height y = 5.00 m at desired velocity v_y, and v_oy = v_o*sin(θ).

We have:

5.00 m = 1.00 m + 21.0 m/s*sin(30.0^o)*t - 1/2*9.81 m/s²*t².

t = 0.496 and t = 1.65, the football is at 5.00 m twice, once on the way up and once on the way down.

Now for the velocity v_y:

v_y = 21.0 m/s sin(30.0^o) - 9.81*0.496 = 5.64 m/s, on the way up

v_y = 21.0 m/s sin(30.0^o) - 9.81*1.65 = -5.64 m/s, on the way down

Have fun! I hope this helps, Joe.