A baseball pitcher throws a fastball horizontally at a speed of 43.0 m/s. Ignoring air resistance, how far does the ball drop between the pitcher’s mound and home plate, 60 \rm ft 6 \rm in away?
Express your answer using SI units.

Question

A baseball pitcher throws a fastball horizontally at a speed of 43.0 m/s. Ignoring air resistance, how far does the ball drop between the pitcher’s mound and home plate, 60 \rm ft 6 \rm in away?
Express your answer using SI units.

anonymous · Answer

60" 6"?
144 m away?
really?
anyway, gravity is constant at about 9.8 m/s
at 144 m away, a ball pitched at 43 m/s will take about 3.3 seconds
3.3 *9.8 = 32.8 
(remember, I'm rounding the numbers when I type them, not when I calculate them)
32.8 is the downward velocity of the ball (should be -32.8), not the distance moved

The equation for distance travelled due to acceleration is
distance = 1/2G*T^2
so, therefore, 
after 3.3 seconds the ball has dropped by (1/2 9.8)*T*T
54.95 metres

So, finally, the ball has dropped a distance of 55 meters when pitched at 43m/s to a target 144 m away.

I am probably wrong. Don't put this down in a test unless you check it...