Question

1. A ball is thrown with an initial speed of 8.60 m/s at an angle of 38.0° above the horizontal. The ball leaves the person's hand 1.20 m above the ground and experience negligible air resistance.                                                                                                                               -
   1. How far from where the child is standing does the ball hit the ground?

b. What is the speed of the ball as it hits the ground?

Answer 1

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