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? correct answer 8.63m
   2. What is the speed of the ball as it hits the ground? correct answer  9.87m/s show all the work and why

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