b. What is the speed of the ball as it hits the ground?
here,
the initial speed , u = 8.6 m/s
theta = 38 degree
the initial height above the ground , h = 1.2 m
a)
let the time taken to hit the ground be t
for vertical direction
- h = u * sin(theta) * t - 0.5 * g * t^2
- 1.2 = 8.6 * sin(38) * t - 0.5 * 9.81 * t^2
solving for t
t = 1.27 s
the horizontal distance traveled , s = u * cos(theta) * t
s = 8.6 * cos(38) * 1.27 m
s = 8.61 m
b)
the final vertical velocity , vy = u * sin(theta) - g * t
vy = 8.6 * sin(38) - 9.81 * 1.27 m/s
vy = - 7.16 m/s
the final horizontal velocity , vx = u * cos(theta)
vx = 8.6 * cos(38) = 6.78 m/s
the speed of ball when it hits the ground , v = sqrt(vx^2 + vy^2)
v = sqrt(6.78^2 + 7.16^2) = 9.86 m/s
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