Question #4153

A ball is thrown upward from the ground with an initial speed of 18.2 m/s; at the same instant a ball is dropped from rest from a building 12 m high. After how long will the balls be at the same height?

Expert's answer

Let's write the equations of motion for both objects:

1. H1= H_{10} + v_{10}t - gt^{2}/2, H_{10}=0, v_{10 }= 18.2 m/s, g = 9.8 m/s^{2}

2. H2 = H_{20 }+ v_{20}t - gt^{2}/2, H_{20}=12, v_{20 }= 0 m/s, g = 9.8 m/s^{2}

Thus

H1 = H2

v_{10}t - gt^{2}/2 = H_{20 } - gt^{2}/2

v_{10}t = H_{20 }

t = H_{20}/v_{10 }= 12/18.2 = 0.66 s

1. H1= H

2. H2 = H

Thus

H1 = H2

v

v

t = H

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