Question

A ball is thrown at a velocity of 35 m/s at an angle of 42º above the horizontal. It hits a wall 96 m away.

Find the time for the ball to reach the top of its trajectory.

Find the maximum height reached.

Find the total time of flight t_f from the origin until it hits the wall.

Find the height y_f at which it hits the wall.

Answer 1

here,

the initial velocity , u = 35 m/s

theta = 42 degree

the horizontal distance of wall , X = 96 m

the time taken by the ball to reach the maximum height , t = (u * sin(theta)) /g

t = (35 * sin(42)) /9.81 s

t = 2.39 s

the maximum height reached , hm = (u * sin(theta))^2 /2g

hm = (35 * sin(42))^2 /(2*9.81) m

hm = 27.95 m

the total time of flight , t2 = X /(u * cos(theta))

t2 = 96 /( 35 * cos(42)) z

t2 = 3.69 s

the height at which the ball hits wall ,yf = u * sin(theta) * t2 - 0.5 * g * tf^2

yf = 35 * sin(42) * 3.69 - 0.5 * 9.81 * 3.69^2 m

yf = 19.63 m