A stone is thrown vertically upwards with an initial velocity of 14m/s. Find the maximum height reached and the time taken. (Please send me the solution with an attachment)
Answers
Answered by
5
Hey ,sorry there is no attachment.But here's the answer.
v=0 u=14m/s g=10m/ss
Since v2=u2-2as
So S=0-14*14/-2*10
=9.8m
Since v=u+at
so 0=14+10t
t= 14/10
= 1.4s
braindeveloper:
nikiswar06 double the time then it will be total
2*1.4
plz vote
at ur service
Answered by
7
From the equation:
at = V - Vo
We find the time that it takes to reach the maximum height
t = (0 m/s - 14 m/s) /( -9.8 m/s^2) = 1.43 s
Neglecting any loose due to friction with air, the object takes the same time to go upwards as to go downwards, So, after being thrown, the object will take twice the time.
Total time = Time Upwards + Time downwards = 1.43 s + 1.43 s = 2.86 s
at = V - Vo
We find the time that it takes to reach the maximum height
t = (0 m/s - 14 m/s) /( -9.8 m/s^2) = 1.43 s
Neglecting any loose due to friction with air, the object takes the same time to go upwards as to go downwards, So, after being thrown, the object will take twice the time.
Total time = Time Upwards + Time downwards = 1.43 s + 1.43 s = 2.86 s
Similar questions