A ball is thrown straight upward, from a point on a roof 10 m above the ground.

Question

A ball is thrown straight upward, from a point on a roof 10 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 58.5 m/s. Consider all quantities as positive in the upward direction. At 5.97 s later, the velocity of the ball is what? A ball is projected upward at time t = 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 80.5 m/s. Consider all quantities as positive in the upward direction. The velocity of the ball when it is 89 m above the ground is what?

Explanation / Answer

(3)

initial velocity is in vertical direction and accelaration is -g all along.

Initial velocity , u = 58.5 m/s

Accelaration , a = -9.8 m/s2

As , v= u + at

v = 58.5 - 9.8 * 5.97

V = 0 m/s (reaches maximum height)

(4)

Initial velocity u = 80.5 m/s

When ball is 89 m above ground , displacement, S = 89 -90 = -1 m.

accelaration , a= -9.8 m/s2

Applying third equation of motion,

v2 = u2 + 2aS

v2 = (80.5)2 + 2*(-9.8)*( -1)

v2 = 6499. 85

v = -80.6 m/s (downward)

Velocity of the ball is - 80.6 m/s when it is 89 m above the ground.

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