Question

a man is standing at the top of a 60m high tower. He throws a ball vertically upwards with velocity 20m/s. After what time will the ball pass him going downward?How long after it release will the ball reach the ground
plz answer fast​

Answer 1

Answer:

Height of tower =60 m

Initial velocity =20m/s

Final velocity =0

Using, v=u-gt

Or,0=20-10t

Or,10t=20

Or,t=2s

Now, time of ascent =time of descent

Therefore, the ball pass the man af after 4 seconds going downward.

Again using, s=ut+1/2gt^2 we get,

60=20t+1/2×10t^2

Or, 60=20t+5t^2

Or,t^2+4t-12=0

Or,t^2+6t-2t-12=0

Or,t(t+6)-2(t+6)=0

Or,(t+6)(t-2)=0

Either, t=6 which is not possible or, t=2 seconds

Therefore, time for the ball to reach the ground after release is:(4+2)=6 seconds

It may help uh

Answer 2

Answer:-

• Given:-

Height of Tower = 60 m

Initial velocity [u] = 20 m/s

Final velocity [v] = 0 m/s

• To Find:-

Time taken by the ball to pass him going downward = ?

Time after which the ball reaches the ground = ?

• Solution:-

• Case 1:-

The ball is thrown upward,

We know,

$\boxed{v = u - gt}$

here,

v = 0 , u = 20m/s ,g = 10 m/s²

Therefore,

→ $0 = 20 - 10t$

→ $10t = 20$

→ $t = \dfrac{20}{10}$

→ $t = 2 sec$ [When thrown up]

Time of ascent = Time of descent

$\pink{\bigstar}$ Time taken by the body thrown upwards to reach its maximum height [h] is known as time of ascent.

$\pink{\bigstar}$ Time taken by a freely falling body to touch the ground is called as time of descent.

Hence,

The time taken by the ball ,

→ t = 2 + 2

→ t = 4 sec

• Case 2:-

Calculate the time taken by the ball to reach the ground,

We know,

$\boxed{s \: = ut \: + \frac{1}{2} g {t}^{2}}$

→ $60 = 20 \times t + \dfrac{1}{2} 10 t^{2}$

→ $60 = 20 t + 5t^2$

→ $5t^2 + 20t - 60 = 0$

→ $t^2 + 4t - 12 = 0$

→ $t^2 + 6t - 2t - 12 = 0$

→ $t(t + 6) - 2(t + 6) = 0$

→ $(t + 6) (t - 2) = 0$

→ $\bf{t = - 6} \: and \: \bf{t = 2}$

• Time as -6 is not possible

• Therefore, t = 2 will be considered

Therefore,

Time taken by the ball to reach the ground after release will be [4 + 2] = 6 seconds.

$\red{\bigstar}$ Alternative Method:-

For AB :

u = 20 m/s

v = 0 m/s

g = - 10 m/s²

• Apply 1st eqn. of motion:-

$\boxed{v = u + at}$

→ $0 = 20 + (-10)t$

→ $t = 2 sec$

• Apply 3rd eqn. of motion:-

$\boxed{v^2 - u^2 = 2as}$

→ $(0)^2 - (20)^2 = 2(-10)s$

→ $-400 = -20s$

→ s = 20 m

For BD :

Total distance = s + H = 20 + 60 = 80 m

u = 0 m/s

∵ Start's from rest

g = 10 m/s²

• Apply 2nd eqn. of motion:-

→ $\boxed{s = ut + \dfrac{1}{2} a t ^2}$

→ $80 = 5t^2$

→ $t^2 = 16$

→ t' = 4 s

So ,

1 . After what time will the ball pass him going downward ( See AB and BC )

→ t + t = 2 + 2 = 4 s

2 . How long after it release will the ball reach the ground ( See AB and BD )

→ t + t' = 2 + 4 = 6 s

$\red{\bigstar}$ Some Additional Information:-

Equations Of Motion:-

• 1st equation of motion:-

$\sf{\boxed{v = u + at}}$

• 2nd equation of motion:-

$\sf{\boxed{s = ut + \dfrac{1}{2} a t ^2}}$

• 3rd equation of motion:-

$\sf{\boxed{v^2 - u^2 = 2as}}$