Question

$\large\bf{Question:}$
A ball is thrown vertically in the air at 120 m/s. After 3 seconds, another ball is thrown vertically. What velocity must the second ball have to päss the first ball at 100 m from the ground ?​

Answer 1

Answer:

The first ball will be 100m from the ground at:

The first ball will be 100m from the ground at:100=120t-4.9t²

The first ball will be 100m from the ground at:100=120t-4.9t²4.9t²-120t+100=0

The first ball will be 100m from the ground at:100=120t-4.9t²4.9t²-120t+100=0t=23.63 seconds

The first ball will be 100m from the ground at:100=120t-4.9t²4.9t²-120t+100=0t=23.63 secondsThen the second ball has to be at 100m at 23.63–3=20.63 secs. So:

The first ball will be 100m from the ground at:100=120t-4.9t²4.9t²-120t+100=0t=23.63 secondsThen the second ball has to be at 100m at 23.63–3=20.63 secs. So:100=20.63v-4.9(20.63²)

The first ball will be 100m from the ground at:100=120t-4.9t²4.9t²-120t+100=0t=23.63 secondsThen the second ball has to be at 100m at 23.63–3=20.63 secs. So:100=20.63v-4.9(20.63²)=20.63v-2,085.42481‬

The first ball will be 100m from the ground at:100=120t-4.9t²4.9t²-120t+100=0t=23.63 secondsThen the second ball has to be at 100m at 23.63–3=20.63 secs. So:100=20.63v-4.9(20.63²)=20.63v-2,085.42481‬20.63v=2,185.42481‬

The first ball will be 100m from the ground at:100=120t-4.9t²4.9t²-120t+100=0t=23.63 secondsThen the second ball has to be at 100m at 23.63–3=20.63 secs. So:100=20.63v-4.9(20.63²)=20.63v-2,085.42481‬20.63v=2,185.42481‬v=105.9343 m/s as the initial velocity of the 2nd ball …………….

Hope it helps.

Answer 2

Explanation:

The first ball will be 100m from the ground at:

⇒ 100 = 120t - 4.9t²

⇒ 4.9t²-120t+100 = 0

⇒ t = 23.63 seconds

Then the second ball has to be at 100m at 23.63–3 = 20.63 secs.

So:

⇒ 100 = 20.63v-4.9(20.63²)

⇒ =20.63v-2,085.42481

⇒ 20.63v = 2,185.42481‬

⇒ v = 105.9343 m/s as the initial velocity of the 2nd ball.