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 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to:____________A. 9.0 sB. 9.4 sC. 9.7 sD. 8.7 sE. 10 s

Answer:

B. 9.4 s

Explanation:

In order to calculate the total time taken by the ball to hit the ground, we first analyze the upward motion. We will use subscript 1 for upward motion. Now, using 1st equation of motion:

Vf₁ = Vi₁ + gt₁

where,

Vf, = Final Velocity in upward motion = 0 m/s (ball stops at highest point)

Vi = Initial Velocity in upward motion = 36.2 m/s

g = - 9.8 m/s² (negative due to upward motion)

t₁ = Time taken in upward motion = ?

Therefore,

0 m/s = 36.2 m/s + (-9.8 m/s²)(t₁)

t₁ = (36.2 m/s)/(9.8 m/s²)

t₁ = 3.7 s

Now, using 2nd equation of motion:

h₁ = (Vi₁)(t₁) + (0.5)(g)(t₁)²

where,

h₁ = distance from top of building to highest point ball reaches = ?

Therefore,

h₁ = (36.2 m/s)(3.7 s) + (0.5)(-9.8 m/s²)(3.7 s)²

h₁ = 133.58 - 66.86 m

h₁ = 66.72 m

No, considering downward motion and using subscript 2, for it.

Using 2nd equation of motion:

h₂ = (Vi₂)(t₂) + (0.5)(g)(t₂)²

where,

h₂ = height of the highest point from ground = h₁ + height of building

h₂ = 66.72 m + 90 m = 156.72 m

Vi₂ = Initial Speed during downward motion = 0 m/s (ball stops for a moment at highest point)

t₂ = Time Taken in downward motion = ?

g = 9.8 m/s²

Therefore,

156.72 m = (0 m/s)(t₂) + (0.5)(9.8 m/s²)(t₂)²

t₂² = (156.72 m)/(4.9 m/s²)

t₂ = √31.98 s²

t₂ = 5.7 s

Now, the total time taken by ball to reach the ground is"

Total Time = T = t₁ + t₂

T = 3.7 s + 5.7 s

T = 9.4 s

Therefore, the correct answer is:

B. 9.4 s

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