A body is thrown vertically upwards with speed of 50ms-2.Find the time taken to return to the ground
Answers
Answer:
Applying 1st law of motion,
v = u + at
Since, body is decelerating with g, substituting it in the formula - ( considering g as 10m/s2)
0 = 50 -10t
t = 5 secs.
During this it will travel - as per 2nd law of motion
s = ut + 1/2 at2
s = 50*5 - 1/2 * 10* 5* 5
s = 125 mtrs
Now during descent it will again travel 125 mtrs and it will be accelerating with g, so again substituting
125 = 0*t + 1/2 * 10* t* t
125 = 5* t* t
Gives us t = 5 secs.
Combining both, we get total time = 10 secs.
Short answer - It takes 5 secs to ascend and will take same time to descend.
This is considering pure motion and non existence of air bouyancy - in which case the time will be more.
Answer:
let u be the speed of body,
u=50m/s
time taken by it to reach ground =time of flight
=time of ascent+time of descent
=u/g+u/g
=2u/g
=2×50/10 (take g=10m/s²)
=10 sec
Explanation: