A blue ball is thrown upward with an initial speed of 22.6 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 22.6 m/s, from a height of 0.8 meters above the ground 2.8 seconds after the blue ball, is thrown, a red ball thrown down with an initial speed of 11.2 m/s from a height of 28.8 meters above the ground. The force of gravity due to the earth results m the balls each having a constant downward acceleration of 9.81 m/s^2. What is the speed of the blue ball when it reaches its maximum height? ______m/s How long does it take the blue ball to reach its maximum height? ____________s What is the maximum height the blue ball reaches? _______m What is the height of the red ball 3.64 seconds after the blue ball is thrown? ________ m How long after the blue ball is thrown are the two balls in the air at the same height? __________ s Which plot correctly shows the velocity of the two balls as a function of time? A B C

Explanation / Answer

here,

initial speed of blue ball , u1 = 22.6 m/s

h1 = 0.8 m

a)

the speed of blue ball at maximum height is zero

b)

the time taken to reach maximum height , t1 = u1/g = 22.6/9.8 = 2.31 s

c)

the maximum height of the blue ball , hmax = u1^2 /( 2g) = 22.6^2 /( 2 *9.8)

hmax = 26.1 m

d)

h2 = 28.8 m

u2 = 11.2 m/s

3.64 s after the blue ball thrown

so, the time taken by the red ball , t2 = 3.64 - 2.8 s = 0.84 s

the height of the red ball be h

(h2 - h) = u2 * t2 + 0.5 * g * t2^2

( 28.8 - h) = 11.2 * 0.84 + 0.5 * 9.8 * 0.84^2

h = 15.9 m

the height of the red ball is 15.9 m

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