Question

a body is projected vertically upwards with a velocity 49m/s. find
a. maximum height
b. time of ascent
c.time of flight
d. velocity when it is at a height of 60m
(g=9.8m/s^2)

Answer 1

Answer:

height = 122.5 m

Time = 5 sec

Total time = 10 sec

Answer 2

Explanation:

Initial velocity , u = 49 m/s

Acceleration due to gravity, a = - 9.8 m/s^2

At maximum height , the final velocity , v = 0 m/s

Using 3rd Equation of Motion,

${v}^{2} - {u}^{2} = 2as$

$s = \frac{ {v}^{2} - {u}^{2} }{2a}$

$s = \frac{ - 2401}{ - 19.6} = 122.5 \: m$

let time of ascent be = t

Using 1st Equation of Motion,

$v = u + at$

$t = \frac{v - u}{a} = \frac{ - 49}{ - 9.8} = 5 \: s$

Time of flight = Time of ascent + Time of descent

But, Time of ascent = Time of descent = 5 s

Therefore, Time of flight = 5 + 5 = 10 seconds.

At s = 60 m

u = 0 m/s

a = 9.8 m/s^2

Again , using 3rd Equation of Motion,

${v}^{2} = {u}^{2} + 2as$

${v}^{2} = 2(9.8)(60) = 1176$

$v = 34.29 \: m {s}^{ - 1}$

Therefore, these are your answers.

I hope it helps you. If you have any doubts, then don't hesitate to ask.