How High Does a Projectile Reach When Fired Upwards?

[Kev1n]

1. An object of mass 1.2kg is fired vertically upwards out of a barrel by a force of 50N acting over a period of 0.75 seconds. I need to check calculations for Velocity of the projectile when it leaves the barrell and the height reached



2. acceleration (a) = F/M, Velocity (v) = Vo + at, Height (h) = Vxt



3. a= 50/1.2 = 41.7ms, therefore v= Vo+at, v = 0+41.7x0.75 = 31.3[ms
Next Height
h=vxt = 31.3 x 0.75 = 23.5m/b]

 

[Andrew Mason]

1. An object of mass 1.2kg is fired vertically upwards out of a barrel by a force of 50N acting over a period of 0.75 seconds. I need to check calculations for Velocity of the projectile when it leaves the barrell and the height reached
2. acceleration (a) = F/M, Velocity (v) = Vo + at, Height (h) = Vxt
3. a= 50/1.2 = 41.7ms, therefore v= Vo+at, v = 0+41.7x0.75 = 31.3[ms
Next Height
h=vxt = 31.3 x 0.75 = 23.5m/b]

Your approach to calculating of v is basically correct. You are essentially using:

m\Delta v = F\Delta t

But you are not factoring in the force of gravity acting in the opposite direction. How much of that 50N force simply balances gravity? So what is the NET force acting on the object? That is the force you must use in the above equation.

In the calculation of height your approach is wrong. .75 seconds is the duration of the force - ie. the time it spends in the barrel. What you are asked to find is the height that the object reaches after leaving the barrel ie after the applied force has ended.

To calculate this use: v = v0 - gt (or the equivalent: a = g = \Delta v/\Delta t AND h = v0t - gt^2/2

Can you figure out what t is? (the time at which it reaches maximum height?).

AM