The solution is as follows. First find the final velocity. Second use the final velocity to find the time.
(final velocity squared) = (initial velocity squared) + 2gh
(vf)^2=(vi)^2+2gh
(vf)^2=(30ft/s)^2+2(32ft/s^2)(80ft)
(vf)^2=900(ft/s)^2+5120(ft/s)^2=6028(ft/s)^2
taking the square root yields
vf=77.59(ft/s)
The final velocity and the time are related by the equation vf=vi+gt, rearranging
t=(vf-vi)/g
t=[77.59(ft/s)-30(ft/s)]/32(ft/s^2=47.59(ft/s)/32ft/s^2=1.49s You can check the answer as follows:
h=vi(t)+(1/2)gt^2
h=30(ft/s)(1.49s)+(1/2)(32(ft/s^2)(1.49s)^2
h=44.7ft+16(ft/s^2)[2.22s^2)
h=44.7ft+35.7ft=80.2ft
which is equivalent to the 80 ft in the original problem except for rounding errors. Note: the notation here is awkward but I haven't found an equation editor iw Wiki.
h^2 means h squared
ft/s^2 means feet per second squared
No. What counts in this case is the vertical component of the velocity, and the initial vertical velocity is zero, one way or another.
The height, in feet, above the ground at time t, H(t) = 40 + 32*t - 16*t2
Get the value of initial velocity. Get the angle of projection. Break initial velocity into components along x and y axis. Apply the equation of motion .
At any time 't' seconds after the ball is released,until it hits the ground,h = 5 + 48 t - 16.1 t2
4ft*Ns=H
initial velocity of the kick = 28.06 m/s
To answer this question one would need to know the rock's initial height and velocity.
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)
No. What counts in this case is the vertical component of the velocity, and the initial vertical velocity is zero, one way or another.
height=acceletation(t^2) + velocity(t) + initial height take (T final - T initial) /2 and place it in for time and there you go
s=u*t +1/2*a*t*t u=32 t=1 a=33 s=48.5 hence height above ground=128-48.5 =79.5 feet
The answer depends on its initial velocity and the height from which its fall to the ground is measured.
4h
39 m\s downward
If you know the initial height and the length of the pendulum, then you have no use for the mass or the velocity. You already have the radius of a circle, and an arc for which you know the height of both ends. You can easily calculate the arc-length from these. And by the way . . . it'll be the same regardless of the mass or the max velocity. They don't matter.
The height, in feet, above the ground at time t, H(t) = 40 + 32*t - 16*t2