Question

A body rises vertically up to height of 225 m in 15 s and then comes back at the point of projection.
find:
a. total distance travelled
b. the displacement
c. the average speed
d. the average velocity

Answer 1

Answer:

A. The total distance travelled will be

225 + 225 = 450m

B. The displacement will be 0 as the body returns to it's initial point.

C. Average speed will be ,

$\frac{450}{30} = 15 \: m {s}^{ - 1}$

225 metres and 15 secondd upwards and 225 metres and 15 seconds downwards. So total 450 metres and 30 seconds.

D. The average velocity will be 0 as displacement is 0.

Answer 2

Answer:

distance =450m displacement=0m avg speed =15m/s avg velocity =0m/s

Explanation:

as distance is total path length while disp is zero cause particle return to original position

avg speed =total distance/totaltime