Help: A baseball leaves the hand of a pitcher 6 vertical feet above home plate a

Question

Help:

A baseball leaves the hand of a pitcher 6 vertical feet above home plate and 60 feet from home plate. Assume that coordinate axes are oriented as shown in the figure. In the absence of all forces except gravity, assume that a pitch is thrown with an initial velocity of ft/s (about 90 mi/hr). How far above the ground is the ball when it crosses home plate and how long does it take for the pitch to arrive? What vertical velocity component should the pitcher use so that the pitch crosses home plate exactly 3 feet above the ground.

Explanation / Answer

a) Velocity is <130, 0, -3>

Component of vel in x axix = 130 ft/s

Distance covered = 60 ft

So time taken = 60/130 = 0.461 secs

Vertical component of vel = -3 ft/s

a = -9.8 m/s2 or -32.15 ft/s2

We know, S = ut + 0.5at^2

Putting that equation, where u = 3, t = 0.461, a = 132.15

S = 4.8 ft

So when the ball crosses home plate, it is (6-4.8) = 1.2 ft above ground

Time take n = 0.461 secs

b) If S = 3ft, a = 32.15, t = 0.461 then u = -0.2 ft/sec

Thus for ball to cross home plate 3 ft above ground vertical velocity should be 0.2 ft/sec upwards.

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