Question

9.What is the distance travelled by a body thrown upward with a speed 20 m/s under the effect of gravity in the first second of its motion.(given g=10m/s2) and also find its velocity at first second.

Answer 1

Answer:

velocity after first second :

v= 20 -10×1

= 10m/s

distance travelled in first second

10^2 - 20^2 = 2 (-10) S

- 300 = -20S

S = 15m

Answer 2

Correct Question:-

What is the distance travelled by a body thrown upward with a speed 20 m/s under the effect of gravity in the first second of its motion.(given g=10m/s2) and also find the distance covered at first second.

Given:-

• A body is thrown upward with a speed 20 m/s under the effect of gravity in the first second of its motion.
• Acceleration due to gravity is 10 m/s².

To Find:-

Distance covered by the object at the first second.

SoLuTioN:-

A body is thrown upward with a speed 20 m/s under the effect of gravity in the first second of its motion. Means, the initial velocity of the body i.e. u is 20 m/s.

Also, the acceleration due to gravity i.e. g is 10 m/s². But as the body is thrown against the gravity. So, acceleration due to gravity is -10 m/s².

We have to find the distance covered by the object at the first second. We can say that given one is 1 second.

At the highest point, final velocity i.e. v will be 0 m/s.

Using the Second Equation Of Motion,

$\huge{\sf{s=ut+1/2at^2}}$

Substitute the known in the above formula,

$\implies\:\sf{s=(20)(1)+1/2(-10)(1)^2}$

$\implies\:\sf{s=20+1/2(-10)(1)}$

$\implies\:\sf{s=20+(-5)(1)}$

$\implies\:\sf{s=20-5}$

$\implies\:\sf{s=15}$

Therefore, the distance covered by the body is 15 m.