A blue ball is thrown upward with an initial speed of 20.8 m/s, from a height of

Question

A blue ball is thrown upward with an initial speed of 20.8 m/s, from a height of 0.6 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 10.5 m/s from a height of 24.6 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/s^2.

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 3.25 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

Explanation / Answer

1) At maximum height velocity will be 'zero'

2) v =u+at
here v=0, at maximum height
u=20.8 m/s
a=-g since it is moving aginst hte gravity
a=-9.81 m/s2

t=20.8/9.81 = 2.12 s

3)h=ut+1/2at2

h=20.8*2.12 +1/2*(-9.81)*(2.12)2

h=22.04 m

4)red ball is thrown aafter 2.5 sec blue ball is thrown

after 3.25 sec blue ball is thrown t of red ball is t=0.75.

u=10.5

s=(10.5*0.75)+1/2*(9.81)*(0.75)2

=7.875 +2.75

=10.63m

h=24.6-10.6= 14 m

5)equations of motion

for blue ball is h=20.8t-1/2 *(9.81)*t2

for red ball is h=24.6-(10.5*t+1/2*(9.81)*t2)

therefore 20.8t-1/2*(9.81)*t2 = 24.6-(10.5*t+1/2(9.81)*t2)

9.81t2 -31.3t+24.6 =0

therefore t=1.4 sec

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