Question

A body is thrown at some angle from the ground.
The magnitude of change in velocity of body in
5 s, if it is still in air, is (g = 10 m/s2]
(1) 50 m/s
(2) 25 m/s
(3) Zero
(4) Data is insufficient​

Answer 1

Answer:

(1) 50 m/s

Explanation:

g=v-u/t

10=v-0/5

: u=0, because the object starts from rest position

v=10*5

v=50 m/s

Answer 2

Answer:

(1)50 m/s

Explanation:

We are given that a body is thrown at some angle from the ground

Time taken by the particle = 5 s

Acceleration of particle =$10m/s^2$

We have to find the magnitude of the body

Let v be the velocity of body when body is still in air  

Initial velocity of particle when particle at rest=u=0

We know that acceleration =$\frac{v-u}{t}$

10 =$\frac{\mid {v-0}\mid}{5}$

Magnitude of change in velocity of body =$10\times 5$

Magnitude of change in velocity of body =50 m/s

Hence, the magnitude of velocity = 50 m /s