Question

9.25 A body is projected vertically upward
from the ground with a velocity of 29.4m/s. Calculate
maximum height attained by the body
and time required to
attain maximum height (g=9.8m/S2)​

Answer 1

Given Information,

• Initial Velocity (u) = 29.4 m/s
• Final Velocity (v) = 0 m/s [Velocity at maximum height is zero.]
• Acceleration due to gravity = - 9.8 m/s² [According to Sign Convention]

Using the third kinematic equation,

v² - u² = 2gh

$\longrightarrow$ 0² - (29.4)² = 2(-9.8)h

$\longrightarrow$ 29.4 × 29.4 = 19.6h

$\longrightarrow$ 3 × 29.4 = 2h

$\longrightarrow$ h = 1.5 × 29.4

$\longrightarrow$ h = 44.1 m

Time of ascent :

t = u/g

$\longrightarrow$ t = 29.4/(-9.8)

$\longrightarrow$ t = |-3|

$\longrightarrow$ t = 3s

The body reaches 44.1 metres in 3 seconds.

Answer 2

Given,

• Initial velocity(u) = 29.4 m/s
• Final velocity(v) = 0 m/s
• Acceleration(g) = 9.8 m/s² (As the ball is projected vertically upward, value of gravitational acceleration is considered)

To find,

• Height attained by the body = ?
• Time required to attain the height = ?

Formula used,

1. v² - u² = 2gs
2. t = u/g

Solution,

Firstly let's find out the maximum height reached by the body when projected vertically upwards.

v² - u² = 2gs

0² - (29.4)² = 2 × 9.8 × s

0 - 846.36 = 19.6s

s = 846.36/19.6

s = 44.1 m

Hence, The maximum height maximum height reached by the body when projected vertically upwards is 44.1 m.

Secondly, let's find out the time required to attain maximum height

t = u/g

t = 29.4 / 9.8

t = 3 s

Hence, The time required to attain maximum height is 3 seconds.