Question

if a ball is thrown up from a tower with an initial velocity of 20 metre per second and it falls on ground in a total time of 7 second find the height of Tower take gravity equal to 10 metre per second​

Answer 1

Question:-

If a ball is thrown up from a tower with an initial velocity of 20m/s. It reaches ground in 7 sec then find the height of the tower.(g = 10 m/s²)

Given:-

• The initial velocity of ball is 20m/s. The time taken by the ball to reach ground is 7 sec.

To Find:-

• Find the height of the tower.

✠ Let the height of the tower be "h"

✰ As we know that : -

✯ initial velocity ( u ) = 20m/s

✯ Time taken ( t ) = 7 sec

✯ gravitational force ( g ) = 10m/s²

️ ➭ $\text{v \: = \: u \: + \: gt \: }$

️ ➭ $\text{v \: = \: 20 \: + \: 10 \: ( \: 7 \: ) \: }$

️ ➭ $\text{v \: = \: 20 \: + \: 70 \: }$

️ ➭ $\boxed{\text{ v \: = \: 90m/s \: }}$

Now,

️ ️➭ $\sf s = ut + 1/2 at^2$

️ ➭ $\sf s = 20(7) + 1/2 \times 10 \times 7^2$

️ ➭ $\sf 140 + 245$

️ ➭ $\sf \boxed{385\:m}$

Answer 2

Total time taken (t) = 7 s

$\vec{u}$ = 20 m/s

Acceleration due to gravity (g) = + 10 m/s² (down)

We know,

$\boxed{ h = ut ± \frac{1}{2} gt^2 }$

⇒ $h = (20 ms^{-1})(7 s) + \frac{1}{2} (10 ms^{-2}) (7s)^2$

⇒ $h = 140 + 5 \times 49$ m

⇒ $h = 140 + 245$ m

⇒ ${h = 385 }$ m (answer).

More:-

Equations of motion influenced by acceleration due to gravity and height obtained:-

1. $v = u ± gt$
2. $h = ut ± \frac{1}{2} gt^2$
3. $± 2gh = v^2 - u^2$.