Question

When a ball is thrown vertically upward it goes through a distance of 19.6 M .Find the initial velocity of the ball and the time taken by it to rise to the highest. Take g as 10 metre per second square.


Find it if you can.✌✌​

Answer 1

Answer:

Initial velocity = 19.8 m/s

time taken = 1.98 s

Explanation:

A ball thrown vertically upwards and the total height covered by it is 19.6 m, it says :

• At the height of 19.6 m, speed of ball at this point is 0.
• Total height covered by that ball is 19.6 m.

Let :

Initial velocity be u.

Total time taken to reach the highest point ( at the height of 19.6 m ) be t.

Using the equations of motion :

At highest point,

= > v^2 = u^2 + 2aS

= > 0 = u^2 + 2( - g ) S { speed at highest point = 0, acceleration is opposite gravity }

= > 0 = u^2 - 2g( 19.6 m )

= > u^2 = 2( 10 m / s^2 )( 19.6 m )

= > u^2 = 392 (m/s)^2

= > u ≈ 19.8 m/s = initial velocity of the ball

= > v = u + at

= > 0 = 19.8 m/s - gt

= > 19.8 m/s = 10(m/s^2) t

= > 1.98 s = t

Initial velocity = 19.8 m/s

time taken = 1.98 s

Answer 2

Answer:

let initial velocity = u m/s

time taken to rise to highest point = t second

final velocity at highest point = 0

g = -9.8 m/s²

distance travelled = 19.6 m

d = (v² - u²)/2a

⇒ 19.6 = (0 - u²)/2×(-9.8)

⇒ 19.6 = u²/19.6

⇒ u² = 19.6×19.6

⇒ u = 19.6 m/s

v = u+at

⇒ 0 = 19.6 - 9.8t

⇒ 9.8t = 19.6

⇒ t = 19.6/9.8

⇒ t = 2s

Initial velocity = 19.6 m/s

time taken to reach highest point = 2s