Question

A ball is thrown with an initial speed of 20 m/s at an angle of 60° to the ground.
If air resistance is negligable, what is the ball’s speed at the instant it reaches its maximum height from the ground?

Answer 1

Given that Rocket is fired at V0 = 20 m/sec at 60 deg above the horizontal

V0x = Initial horizontal velocity = V0*cos = 20*cos 60 deg = 10 m/s

V0y = Initial vertical velocity = V0*sin = 20*sin 60 deg = 17.32 m/s

Since there is no acceleration in horizontal direction so it's horizontal velocity will remain constant, So at max height

Vfx = V0x = 10 m/s

At max vertical velocity of projectile becomes zero, So

Vfy = 0 m/s

So,

Vf = (10 i + 0 j)

Speed of projectile will be:

|Vf| = sqrt (10^2 + 0^2)

|Vf| = 10 m/s = speed of ball at max height