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

$\huge\bold{Answer}$

Give 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 kinetic equation,

v² - u² = 2gh

⟶ 0² - (29.4)² = 2(-9.8)h

⟶ 29.4 × 29.4 = 19.6h

⟶ 3 × 29.4 = 2h

⟶ 1.5 × 29.4

⟶ h = 44.1 m

Time of ascent :

t = u/g

⟶ t = 29.4/(-9.8)

⟶ t = l-3l

⟶ t = 3s

The body reaches 44.1 metres in 3 seconds.

Hope it's helpful :)

Answer 2

We have,

a = g = - 10 m/s² (up)

u = 29.4 m/s

v = 0 m/s (will remain rest)

We know,

$2g∆h = {v_f}^2 - {v_i}^2$

⇒ $∆h = \dfrac{{v_i}^2}{2g}$

⇒ $∆h = \dfrac{(29.4 \ ms^{-1})^2}{2 \times 9.8 \ ms^{-2}}$

⇒ $∆h = 44.1 m$.

Now we know,

$v_f = v_i + gt$

⇒ $v_i = gt$

⇒ $t = \dfrac{v_i}{g}$

⇒ $t = \dfrac{29.4 \ ms^{-1}}{9.8 \ ms^{-2}}$

⇒ $t = 3 s$.

∴ It will cover 44.1 m height and will take 3 seconds to reach the apex.