Question

A baseball, hit 7 feet above the ground, leaves the bat at an angle of 55 ° and is caught by an outfielder 6 feet above the ground and 48 feet from home plate. Neglecting air resistance, what was the initial speed of the ball? Use g = 32 ft/sec2 as the acceleration of gravity. Your answer should be in feet per second.

Answer 1

Answer:

The answer would be $V_{0}=40,429 ft/s$

Explanation:

In this problem, we will use the range that is given and the equation of the flying time.

The ball leaves the bat at an angle so the speed of the ball has x and y vectors. So as i draw in the picture the speed has $V_{x} =Vcos55$ and $V_{y} =Vsin55$

Also $V_{x}$ speed is constant through flying time

So to find the range we use the equation:

$X_{range} =V_{x} t=Vcos55t$

Now, we know the range but to find V, we need the time t.

Total flying time can be found using the equation:

$t_{flying} =\frac{2V_{y} }{g}$

If we put t in the first equation:

$48=Vcos55\frac{2Vsin55}{g}$

Now since we also know g if we solve the equation we find $V=40,429ft/s$