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 Question from Kathy, a teacher: If the angle at time 0 is not 45 degrees how can you find the initial velocity? Ball thrown from height of 7 feet. Caught by receiver at height of 4 feet after traveling 90 feet down the field. Find initial velocity. You had a similar problem answered but the angle was 45 degrees so the cos and sin were equal and the equations were simpler to work with. Thank you!

Hi Kathy,

There is not enough information to determine the initial velocity. The initial velocity depends on the angle.

The parametric equations given in the earlier question are

$x(t) = (v_0 \cos \theta)t \mbox{ and } y(t) = h + (v_0 \sin\theta)t-16t^2.$

If $\theta$ is 45 degrees then using the data given and letting $t_1$ be the time when the ball is caught these become

$x(t) = \left(v_0 \times \frac{1}{\sqrt{2}}\right)t_1 = 90 \mbox{ and } y(t) = 7 + \left(v_0 \times \large \frac{1}{\sqrt{2}} \right)t_1 -16t_1^2 = 4 .$

As I indicated in my earlier response you can substitute $(v_0 \times \frac{1}{\sqrt{2}})t_1 = 90$ into the second equation and solve for $t_1$ then substitute the value of $t_1$ into $(v_0 \times \frac{1}{\sqrt{2}})t_1 = 90$ and solve for $v_0.$ I got $t_1$ to be a little less than 2.5 seconds and $v_0$ about 53 feet per second. (You should check my arithmetic.)

What if $\theta$ is 30 degrees? This would give

$x(t) = \left(v_0 \times \frac{\sqrt{3}}{2} \right)t_1 = 90 \mbox{ and } y(t) = 7 +\left (v_0 \times \large \frac{1}{2} \right)t_1 -16t_1^2 = 4 .$

This time substituting the first equation into the second I got $16 t_1^2 = \normalsize \frac{90}{\sqrt{3}} + 3$ which gave me $t_1$ approximately 1.9 seconds and $v_0$ almost 107 feet per second.

Throwing at a lower arc means the ball gets to the receiver more quickly but you need to throw it much harder.

I hope this helps,
Harley

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.