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I am working on a program that shoots a projectile straight up into the air. I need to be able to calculate the initial velocity required to reach a height at a given amount of seconds after firing.

I know the initial velocity ignoring setting the travel time is:

$v_0 = \sqrt{2gh}$

where

$g$ - gravity - could be other than $9.81m/s^2$

$h$ - maximum height to reach

$t$ - time to reach height, already known

But how do I calculate the continious downward force needed, so that after $t$ amount of seconds the object reached $h$, therefore has velocity of 0, then falling down and reach the ground again after $t$ seconds ?

The smaller I set $t$, more initial velocity is required and downward force gets also greater, am I right ?

Thanks in advance.

Tobi123
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Mc Midas
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1 Answers1

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You do not need to consider downward force explicitly. It is enough to know that the deceleration due to gravity is always -g. Thus, the time taken to reach zero velocity is simply obtained from the deceleration equation below,

$$0-v_0=-gt$$

which leads to

$$t=\frac{v_0}{g}$$

So, the larger the initial velocity is, the longer it takes to reach zero velocity.

If you were able to change the gravity (downward force), the new gravity $G$ required for the projectile to reach $h$ in $t$ would be

$$G=\frac{2h}{t^2}$$

Quanto
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  • Your right, but I already have defined the time it takes to reach h, so v0 should be greater to move the object faster to h, but it should not exceed h. – Mc Midas Aug 10 '19 at 00:38
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    Right, but t and h can not be specified independently. One determines the other according to the last equation. – Quanto Aug 10 '19 at 00:45
  • So to move it faster to h, I must increase v0, but due that it takes more time to reach zero velocity I must modify the downward force so that it applies enough force to stop the object at h. -g would be not enough. – Mc Midas Aug 10 '19 at 11:59
  • To add, I would like to have the constant that increases v0 defined as the travel time, so less travel time, more v0. But as you said t and h cannot be set independently. So is this impossible? – Mc Midas Aug 10 '19 at 12:07
  • It’d be possible if you could change $g$. – Quanto Aug 10 '19 at 23:13