Question

A ball is thrown vertically upward with a velocity of 49 M per second calculate
A. The maximum height to which it Rises
B. The total time it takes to return to the surface of the earth

Answer 1

3rd law for newton ,

Vt² = V₀² + 2 a x

Vt = 0 for max. height

V₀ = 49 m/s

a = g = 9.81 m/s² ( - upward )

x = the height req.

so x = 122.375 m

1st law ,

Vt = V₀ + a . t

0 = 49 + (-9.81).t

t ( time req. to reach max height ) = 5 sec

the time to return to the earth = 2t 10 sec

Answer 2

u=49m/s.

v=0m/s.

g=-9.8m/s.

s=h= ?.

¡) from equation of motion.

2as= v square- u square.

2×(-9.8)×h=0 square- 49 square.

h= 112.8m.

¡¡) from equation of motion.

v=u+at.

t=v-u/a.

t=0-49/9.8.

=5s.

Total time taken by ball to reach back to the ground = 5+5 = 10s.