Asked by chibest

A batted baseball leaves the bat at an angle of 33.0∘ above the horizontal and is caught by an outfielder 380ft from home plate at the same height from which it left the bat.

1. What was the initial speed of the ball?
2. How high does the ball rise above the point where it struck the bat?
Answered by bobpursley
time in air:

horizontal distance=speed*time
380=Vcos33*t
t=380/Vcos33

Vertical
hf=hi+vi*t-4.9 t^2
hf, hi are zero...

t(Vsin33-4.9t)=0

Vsin33-4.9(380/Vcos33)=0
V^2=4.9*380/sin33cos33
solve for V

how high, now solve for t/2,
h=vi*t-1/2 g t^2 where t is half the time.
Answered by chibest
A snowball rolls off a barn roof that slopes downward at an angle of α= 44.0∘ . (See the figure below) The edge of the roof is H= 17.0m above the ground, and the snowball has a speed of v = 7.00m/s as it rolls off the roof. Ignore air resistance

1. How far from the edge of the barn does the snowball strike the ground if it doesn't strike anything else while falling?
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