According to a friend standing on the ground next to the track, what is the ball

Question

According to a friend standing on the ground next to the track, what is the ball's initial speed m/s What is the angle of the launch degree above the horizontal What is the displacement of the ball during its rise x-component: m y-component: m A projectile is launched from ground level with an initial speed of 54.7 m/s. Find the launch angle (the angle the initial velocity vector is above the horizontal) such that the maximum height of the projectile is equal to its horizontal range. A stone thrown horizontally from the top of a 20-m tower hits the ground at a point 20 m from the base of the tower

Explanation / Answer

3) initial speed , u = 54.7 m/s

let the launch angle is theta

maximum height = horizontal range

v^2 * sin^2(theta)/(2 * g) = 2 * v^2 * sin(theta)*cos(theta)/g

sin(theta)/(2) = 2 * cos(theta)

tan(theta) = 4

theta = 76 degree

for launch angle is 76 degree for maximum height is equal to horizontal range

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