Question

A baseball is hit with an initial speed of 35 m/s at an angle of 37 degrees with respect to the horizontal. How long does it take to hit the ground. You may assume that the batter hits the ball 1 meter above the ground

Answer 1

a)

for vertical motion,using the equation of motion,

s = u*t+0.5*a*t^2

where u = initial (vertical)velocity = 35*sin(37 degrees)

s = v ertical displacement = -1 m <----- ve sign for downward motion

a = -9.8 m/s2

So, -1 = 35*sin(37 degrees)*t + 0.5*(-9.8)*t^2

So, t = 4.35 s <-------------answer

Answer 2

Vo = 35m/s @ 37 deg.

Yo = ver. = 35sin37 = 21.06m/s.

h = (0 - (21.06)^2) / -19.6 = 22.63m

Answer 3

First it will move in a projectile path,

Vertical Velocity = Vsina

Time of Flight (T1)= 2Vsina/g

after that it will just fall freely for a height of 1m, Time (T2) = (2/g)^1/2

Therefore total time = T1 + T2 = 2*35*sin37/(9.8) + (2/9.8)^1/2

= 4.75 sec