a partical is thrown up with velocity 10m/s.fund the following:
a) Height at which the particle turns
b)Time to reach the highest point
c) Velocity of the particle at the initial point
b) Velocity of the at half of the height covered
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Answer:
The initial speed is υ
0
=10. The horizontal component of velocity is
υ
0
cos60
0
=0.5υ
0
=5
At the highest point of the trajectory of the projectile, its speed is 5m/s which is half the initial speed.
The time taken to reach the maximum height (when the vertical component of the speed becomes 0) is
0=υ
0
cos60
0
=−gt
t=
10
10(
2
3
)
t=0.866sec
The required time is t=0.866 sec
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