Question

a ball is thrown upward with a velocity of 35m/s and a ball is dropped from a height of 100m with a velocity of 10m/s. When and where the two balls will meet?​

Answer 1

Answer:

• Two balls will meet after a time of 2.2 seconds.
• And they will meet at a distance of 45.72 metres above the ground.

Explanation:

Given that,

The Ball is thrown upward with a velocity of 35 m/s so,

• Initial velocity of first ball, u₁ = 35 m/s
• Acceleration due to gravity, a₁ = -9.8 m/s²

[ Negative acceleration due to gravity because the ball is going upward ]

Also,

Another ball is dropped from a height of 100 m with a velocity of 10 m/s, so

• Initial velocity of second ball, u₂ = 10 m/s
• acceleration due to gravity, a = 9.8 m/s²

We need to find,

• When and where the two balls will meet

So,

Let, first and the second ball meet after a time 't' and covering s₁ and s₂ distances respectively

then,

For the first ball thrown upward, Using the second equation of motion

→ s₁ = u₁t + 1/2 a₁t²

→ s₁ = 35 t + 1/2 (-9.8) t²

→ s₁ = 35 t - 4.9 t²  ____eqaution (1)

Now,

For the second ball thrown downward, Using the second equation of motion

→ s₂ = u₂t + 1/2a₂t²

→ s₂ = 10 t + 1/2 (9.8) t²

→ s₂ = 10 t + 4.9 t²    _____equation (2)

Since one ball is thrown from the ground and another from a 100 m high building, therefore

→ s₁ + s₂ = 100 m

→ 35 t - 4.9 t² + 10 t + 4.9 t² = 100

→ 45 t = 100

→ t = 100 / 45

→ t = 20/9 = 2.2  seconds

putting the value of t in equation (2)

→ s₂ = 10 t + 4.9 t²

→ s₂ = 10 (2.2)  +  4.9 (2.2)²

→ s₂ = 22  +  23.716 = 45.716

→ s₂ = 45.72 metres   [Approx.]

Therefore,

• Two balls will meet after a time of 2.2 seconds.
• And they will meet at a distance of 45.72 metres above the ground.

Answer 2

Answer:

Two balls will meet after a time 10/(3√5) seconds or 1.49 seconds.

And they will meet at a distance of 25.8 metres above the ground.

Explanation:

Given that,

Ball is thrown upward with a velocity of 35 m/s so,

Initial velocity of first ball, u₁ = 35 m/s

Acceleration due to gravity, a₁ = -9.8 m/s²

[ Negative accleration due to gravity because ball is going upward ]

Also,

Another ball is dropped from a height of 100 m with a velocity of 10 m/s,so

Initial velocity of second ball, u₂ = 10 m/s

acceleration due to gravity, a = 9.8 m/s²

We need to find,

When and where the two balls will meet

So,

Let, first and second ball meet after a time 't' and covering s₁ and s₂ distances respectively

then,

For first ball thrown upward, Using second equation of motion

→ s₁ = u₁t + 1/2 a₁t²

→ s₁ = 35 t + 1/2 (-9.8) t²

→ s₁ = 35 t - 4.9 t²  ____eqaution (1)

Now,

For second ball thrown downward, Using second equation of motion

→ s₂ = u₂t + 1/2a₂t²

→ s₂ = 10 t + 1/2 (9.8) t²

→ s₂ = 10 t + 4.9 t²    _____equation (2)

Since, one ball is thrown from ground and another from a 100 m high building, therefore

→ s₁ + s₂ = 100 m

→ 35 t - 4.9 t² + 10 t + 4.9 t² = 100

→ 45 t² = 100

→ t² = 100 / 45

→ t = 10/(3√5)   seconds

putting value of t in equation (2)

→ s₂ = 10 t + 4.9 t²

→ s₂ = 10 (10/(3√5))  +  4.9 (10/(3√5))²

→ s₂ = 100/(3√5)  +  490 / 45

→ s₂ = (100√5) / 15  +  490 / 45

→ s₂ = ( 300 √5 + 490 ) / 45

→ s₂ = 25.8 metres

Therefore,

Two balls will meet after a time 10/(3√5) seconds or 1.49 seconds.

And they will meet at a distance of 25.8 metres above the ground.