For this problem, we will use the formula for position of an object. y=1/2at^2+vot+ho, where t is time in seconds, a is acceleration in meters per second squared (in this case gravity so -9.8 m/s2), v0 is the initial velocity in meters per second, and h0 is the initial height in meters.

Question

A pitcher is standing 20 meters horizontally from a batter. Let’s assume home plate is at the origin. As the pitcher releases the ball it is at a height of 2 meters off the ground and we consider this time t = 0 seconds. He throws the ball horizontally with an initial velocity of 40 meters per second. For the sake of the problem we will assume that the horizontal velocity stays at a constant 40 meters per second.
a. Give the parametric equations for the position of the ball as a function of time.

b. Show that the height of the ball when it reaches the batter is 0.775 meters.

c. If the pitcher throws the ball with an angle of depression of 10 rather than horizontally, what happens by the time it reaches the batter.

Answer 1

(a) let x=t, y=h(t)
(b) as with all parabolas, the vertex is at -b/2a
(c) vo is v sinθ

now crank it out

Answer 2

where does the 1/2 come from?