Question

a body of mass 2kg is dropped from a height of 80m. g = 10 m/s 2then find
1. time taken to reach ground
2. velocity with which it reaches the ground​

Answer 1

Answer:

1. h = ut + 1/2 gt^2

u = 0 m/s

g= 10 m/s^2

h = 1/2 × 10 × t^2

80 × 2 / 10 = t^2

160/ 10 = t ^2

t^2 = 16

t = 4 seconds

2. v = u + gt

v = 10 × 4

v = 40 m/s.

Answer 2

$\dag\:\underline{\sf AnsWer : }$

Here we are given a body whose mass (m) is 2 kg and the initial velocity (u) will be 0 m/s and the body is dropped from height of 80 m i.e, distance (s) = 80 m and acceleration due to gravity (g) = 10 m/s². We are asked to find the time taken by the body to reach ground (t) and velocity with which it reaches the ground (v).

• So, first we will find the time taken by the body to reach ground (t) by using second kinematical equation of motion :

$:\implies \sf s = ut + \dfrac{1}{2} gt^2$

$:\implies \sf 80 = 0 \times t+ \dfrac{1}{2} \times 10 \times t^2$

$:\implies \sf 80 = \dfrac{1}{2} \times 10 \times t^2$

$:\implies \sf 80= 5\times t^2$

$:\implies \sf t^2 = \dfrac{80}{5}$

$:\implies \sf t^2 = 16$

$:\implies \sf t= \sqrt{16}$

$:\implies \underline{ \boxed{\sf t= 4 \: second}}$

• Now, by using third kinematical equation of motion we can find the velocity with which it reaches the ground

$\dashrightarrow\:\:\sf v^2 - u^2 = 2as$

$\dashrightarrow\:\:\sf v^2 - (0)^2 = 2 \times 10 \times 80$

$\dashrightarrow\:\:\sf v^2 = 1600$

$\dashrightarrow\:\:\sf v= \sqrt{1600}$

$\dashrightarrow\:\: \underline{ \boxed{\sf v= 40 \: m/s}}$