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A baseball player hits a baseball that lands in stands 24 meters above the playing field.The ball land with velocity of 50 meters per second at angle of 35 degree below the horizontal.

a.If a player hits the ball 1 meter above the playing field,what was the velocity of the ball upon leaving the bat?

b.What was the horizontal distance traveled by ball

c.how long was the ball in air

I have done like this:

let r be angle in degrees

t=2usinr/g

time=?

velocity=50m/s

gravity=10

time= 2*50*sin35/10=5.74s

velocty of ball at 1m will be:

y=vsinrt-0.5gt

1=usin35*5.74-0.5*10*5.74*5.74

u=50.61m/s

b.horizontal distance traveled by ball

distance/time=velocity

then

distance/time=ucosr

distance=50cos35*5.74

distance=235m

c.Time of the ball in air.

t=2usinr/g

time=?

unitial velocity=50m/s

gravity=10

time= 2*50*sin35/10=5.74s

Any idea about this question?