A projectile is launched vertically upward from ground level with an initial velocity of 112 ft/s.
A. Find the velocity at t= 3s and t=5s
B. How high will the projectile rise
C. Find the speed of the projectile when it hits the ground.
A projectile is launched vertically upward from ground level with an initial velocity of 112 ft/s.
A. Find the velocity at t= 3s and t=5s
B. How high will the projectile rise
C. Find the speed of the projectile when it hits the ground.
What have you tried? Since this is going straight up (and then back down, of course) its acceleration is that of gravity, -g. The acceleration is a constant, -g so its speed, at time t is the integral of that. And the distance will be the integral of the speed. The only integral formula you need is $\displaystyle \int t^n= \frac{1}{n+1}t^{n+1}+ C$.
Now try to finish it.
The equations you have to use are :
1)v=u-gt ,where v is the final speed u the initial and g the constant accelaration of bodies falling in empty space
2) h =$\displaystyle ut-\frac{1}{2}gt^2$ ,where h is the height t is the time
So in (1) put t=3s and t= 5s to find the speed
To find out how high the projectile will rise ,put at the equation(1) v=0 and :
t= u/g .Now put that into (2) and we get :h =$\displaystyle \frac{u^2}{2g}$
Note at the point where the projectile reaches its highest point its v =0
I will let you do part c
Using integration we have : ( Bold letters denote vectors)
dv/dt = g.But g =-zg ,where z is the unit vector along the z axis
And since v = zv we have:
z$\displaystyle \frac {dv}{dt} = $-zg and
dv = -gdt . Integrating the left side from v=u to v=v and the right side from t=0 to t=t we get:
v=u-gt.
Also we have :
dh/dt= v. But h=zh and v=zv.
Hence zdh = z(u-gt)dt => dh =(u-gt)dt.
Now integrating the left side from h=0 to h=h and the right side from t=0 to t=t we have:
h =ut-$\displaystyle \frac{1}{2}gt^2$