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A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m/s. At the same instant another ball B is thrown upward from the ground with an initial velocity of 20 m/s. Determine the height from the ground and the time at which they pass.

Answer :

Explanation:

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The height from the ground and the time at which the balls passes is mathematically given as

yB=19.994m

t=1.750s

What are the height from the ground and the time at which they pass?

Question Parameter(s):

A ball A is thrown vertically upward from the top of a 30-m-high

The initial velocity of 5 m/s.

Generally, the equation for the reference point is the bottom of the building  is mathematically given as

y=y0+v0*t+0.5*a*t^2

Therefore

For Individual ball, it will be

yA=28+4*t-4.9*t^2

yB=0+20*t-4.9*t^2

Where

yA=yB

28+4*t-4.9*t^2=20*t-4.9*t^2-16t=-28

t=1.750s

In conclusion, The value of yA,yB is

y_B=20*1.75-4.9*1.75^2

yB=19.994m

Thereofre

y_A=19.994m

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