A blue ball is thrown upward with an initial speed of 21 m/s, from a height of 0

Question

A blue ball is thrown upward with an initial speed of 21 m/s, from a height of 0.7 meters above the ground. 2.6 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 7.5 m/s from a height of 24.9 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

1)

What is the speed of the blue ball when it reaches its maximum height?

m/s

2)

How long does it take the blue ball to reach its maximum height?

s

3)

What is the maximum height the blue ball reaches?

m

4)

What is the height of the red ball 3.38 seconds after the blue ball is thrown?

m

5)

How long after the blue ball is thrown are the two balls in the air at the same height?

s

6)

A

B

C

7)

Which statement is true about the blue ball after it has reached its maximum height and is falling back down?

The acceleration is positive and it is speeding up

The acceleration is negative and it is speeding up

The acceleration is positive and it is slowing down

The acceleration is negative and it is slowing down

Please I don't want the steps! just write the final answers for each

Explanation / Answer

here,

initial speed of the blue ball , ub = 21 m/s

1)

the speed of blue ball when it reaches the maximum height is 0 m/s

2)

let the time taken to reach the maximum height be t

using first equation of motion

vb = ub - g * t

0 = 21 - 9.8 * t

t = 2.14 s

the time taken to reach the maximum height is 2.14 s

3)

let the maximum height be h

h-h0 = ub * t - 0.5 * g * t^2

h - 0.7 = 21 * 2.14 - 0.5 * 9.8 * 2.14^2

h = 23.2 m

the maximum height reached by blue ball is 23.2 m

4)

time taken by the red ball , t' = 3.38 - 2.6 = 0.78 s

initial height , h0 = 24.9 m

h - h0 = ur*t' - 0.5 * g * t'^2

h - 24.9 = 7.5 * 0.78 - 0.5 * 9.8* 0.78^2

h = 27.77 m

the height reached by the red ball is 27.77 m

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