A red ball is thrown down with an initial speed of 1.3 m/s from a height of 27 m

Question

A red ball is thrown down with an initial speed of 1.3 m/s from a height of 27 meters above the ground. Then, 0.5 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 23.9 m/s, from a height of 0.7 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 red ball right before it hits the ground?
2) How long does it take the red ball to reach the ground?
3) What is the height of the blue ball 2 seconds after the red ball is thrown?

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

Explanation / Answer

1)

for red ball

u1 = -1.3 m/s

a1 = -g = -9.81 m/s^2

y1 = -27 m

v1^2 - u1^2 = 2*a1*y1

v1^2 - 1.3^2 = 2*9.81*27

v1 = 23.05 m/s <<------answer

2)

from

v1 = u1 + g*t1

t1 = (v1-u1)/g

t1 = (23.05-1.3)/9.81

t1 = 2.22 s <<<<--answer

3)

y2 = u2*dt + 0.5*a*dt^2

y2 = 0.7 + (23.9*(2-0.5))-(0.5*9.81*(2-0.5)^2)

y2 = 25.5 m

4)

for red ball

y - 27 = u1*t + 0.5*a*t^2

y = 27 - 1.3t - 4.9*t^2

for blue ball

y = 0.7+ u2*(t-0.5) + 0.5*a*(t-0.5)^2

y = 0.7 + 23.9t - 11.95 - 4.9(t-0.5)^2

therefore

t = 0.767 s <<<--------answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.