Question

An object thrown vertically upwards reaches at maximum height in 5 sec .what is the velocity with which it was thrown


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Answer 1

Answer:

50 m/s

Explanation:

At maximum height, speed of the ball(v) is 0.

Acceleration = g = 10 m/s^2.

Let initial speed be u.

Time taken to reach max. height ( t ) = 5 sec

By applying Newton's 3rd equation of motion for a body moving vertically upwards, we have

u = v + gt

substituting v = 0, g = 10 and t = 5,

we get u = 50 m/s.

If you use 9.8 m/s^2 as acceleration due to gravity, just replace 10 by 9.8 in the above calculations and the answer comes as 49 m/s.

Hope this helps.

Answer 2

Here are the two equations you should use to solve this. First vf=vi +a*t, where vf is final velocity (what’s that at the end of going up?), vi is initial velocity, a= guess what, and t=the thing in question. Substitute the right values and solve for t.

I assume that when you ask how will it rise you do not mean like “blessed is he who sits on a tack, for he shall rise”, but you mean “how high will it rise? “

For that you need this equation: d= t*(vf +vi) /2, where d is the vertical distance, and you should already have the other variables. Now, if you’re still interested, compute!
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