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An object can be thrown vertically upwards or at an angle to the ground, in both cases it is needed that time of flight be 5 seconds. This means it's time of ascent (going up) is 2.5 seconds and time of descent(coming down) is also 2.5 seconds. So, it reaches highest point 2.5 seconds after it is thrown. At highest point the vertical component of velocity1 of the object becomes zero for an instance.

Now, kinematics equation can be used to solve this.

v = uy - g*t

where v is final velocity at top =0. uy is initial vertical velocity1. g is accleration due to gravity(9.8ms-2). t is time of ascent.

putting values we get.

0 = uy - 9.8 * 2.5

or uy = 24.5 ms-1

So we need to throw an object with vertical velocity = 24.5 meters per second so that it remains in air for 5 seconds.

1. If object is thrown at an angle then vertical component of velocity of projection is taken.

If object is thrown vertically upwards then vertical component of velocity of projection is

same as velocity.

vertical component of velocity of projection(uy) = u*sin(θ), where u is velocity of projection

and θ is angle of projection with respect to horizontal.

2. Accleration due to gravity is different for different places. 9.8ms-2 is an approximate

value.

3. Here, air resistance and wind speeds have been neglected as they make the calculation very

tedious and they are always varying from time to time and place to place.

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Q: How fast must an object be thrown up if it is to stay in the air for 5 seconds before hitting the ground again?
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