Question

A body is dropped freely under gravity from the top of a tower of height 78.4 m. Calculte, the(i) the time to reach thee ground and the (ii) the velocity with which is strikes the ground
g = 9.8
help me fast

Answer 1

Answer:

• 4s.
• 39.2 m/s.

Explanation:

Given:

⇒ Initial velocity (u) = 0 m/s

⇒ Height (h) = 78.4 m

⇒ Acceleration (g) = 9.8 m/s²

Solution:

(i) From equation of motion h = ut + 1/2gt²

⇛ 78.4 = 0 × t + 1/2 × 9.8 × t²

⇛ 78.4 = 4.9t²

⇛ t² = 78.4/49 = 16

∴ Time taken to reach the ground t = √16 = 4s.

(ii) From equation of motion v = u + gt

⇛ v = 0 + 9.8 × 4

⇛ v = 39.2 m/s

∴ The body strikes the ground with a velocity 39.2 m/s.

Answer 2

$\bf \leadsto \: given \: :$

A body is dropped freely under gravity from the top of a tower.

• Height of the tower h = 78.4 m
• initial velocity u = 0 [ At rest ]

$\bf \leadsto \: we \: have \: to \: \: find \: :$

(i) the time to reach the ground = t

(ii) the velocity with which is strikes the

ground = v

$\bf \leadsto \: \: solution \: :$

${ \boxed{\boxed{\begin{array}{cc} \bf(i) \\ \rm \: we \: know \: that \\ \\ \rm \: h = ut + \frac{1}{2} g {t}^{2} \\ \\ \rm \implies \: 78.4 = 0 \times t + \frac{1}{2} \times 9.8 \:\times\: {t}^{2} \\ \\ \rm \implies \: {t}^{2} = \frac{78.4 \times 2}{9.8} \\ \\\rm \implies \: {t}^{2} =16 \\ \\ \rm \implies \: t = \sqrt{16} \\ \\ \rm \implies \: t = 4 \: s \\ \\ \pink{\boxed{\rm \therefore \: time \: \: t = 4 \: s}}\end{array}}}}$

${ \boxed{\boxed{\begin{array}{cc}\bf \:(ii) \\ \sf \: we \: know \: that \\ \\ \rm \: {v}^{2} = {u}^{2} + 2gh \\ \\\rm \implies \: {v}^{2} = {0}^{2} + 2 \times 9.8 \times 78.4 \\ \\\rm \implies \: {v}^{2} = 1536.64 \\ \\ \rm \implies \: v = \sqrt{1536.64} \\ \\\rm \implies \: v = 39.2 \: m {s}^{ - 1} \\ \\ \pink{ \boxed{ \therefore \rm \: striking \: velocity \: \: v = 39.2 \: m {s}^{ - 1}}} \end{array}}}}$