i dont understand how to do this problem, please explain with answer and how you got it
a penny is tossed from a 240 ft. building with a initial velocity of 23 ft/sec
what is its position after 2 seconds?
what is its velocity after 2 seconds?
how long until it hits the floor?
a penny is tossed from a 240 ft. building with a initial velocity of 23 ft/sec
what is its position after 2 seconds?
what is its velocity after 2 seconds?
how long until it hits the floor?
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This is a classical problem in differential calculus where you manipulate the position equation: y(t) = yo + vo_y*t + 1/2*g*t^2.
Remember: differentiate position you get velocity; differentiate velocity you get acceleration. you can go the opposite direction by integration, but don't worry about that for now.
Also keep in mind we are using English Units of length (feet) in this problem instead of meters, and the acceleration due to gravity in feet is -32 ft/s^2.
Consider what happens after 2 s.
The position at t = 2s is simply found by substituting what you know into the position equation:
y(2) = 240 + 23*2 + 0.5*(-32)*2^2 = 222 ft
Get the velocity function by taking the derivative of position (you may already recognize this from previous problems and / or physics).
vy(t) = vo_y + g*t
vy(t) = 23 - 32*t
So vy(2) = -39 ft/s, which means the penning is going down at 39 ft/s
For the final part you can use the position equation with the final position being y = 0 (this means the penny hits the floor), and solve the resulting quadratic equation for the time t.
Using the values from parts a & b you get the quadratic equation:
0 = -16*t^2 - 39*t + 222
which you can solve using your favorite method (like the Quadratic Formula) to get t = 2.70s.
Remember: differentiate position you get velocity; differentiate velocity you get acceleration. you can go the opposite direction by integration, but don't worry about that for now.
Also keep in mind we are using English Units of length (feet) in this problem instead of meters, and the acceleration due to gravity in feet is -32 ft/s^2.
Consider what happens after 2 s.
The position at t = 2s is simply found by substituting what you know into the position equation:
y(2) = 240 + 23*2 + 0.5*(-32)*2^2 = 222 ft
Get the velocity function by taking the derivative of position (you may already recognize this from previous problems and / or physics).
vy(t) = vo_y + g*t
vy(t) = 23 - 32*t
So vy(2) = -39 ft/s, which means the penning is going down at 39 ft/s
For the final part you can use the position equation with the final position being y = 0 (this means the penny hits the floor), and solve the resulting quadratic equation for the time t.
Using the values from parts a & b you get the quadratic equation:
0 = -16*t^2 - 39*t + 222
which you can solve using your favorite method (like the Quadratic Formula) to get t = 2.70s.
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I suppose that the motion is vertical
s(t) = (g/2)t² + v₀t
where s=240ft is the space, g is gravity acceleration g ≈ 32.174 ft/s²
v₀=23 ft/s is the initial velocity
Plug data into the formula
s(t) = 16.087t² + 23t
v(t) = ds/dt = 32.174t + 23
after 2 seconds
s = 110.348ft
v = 87.348ft/s
hits the floor when s = 240ft
16.087t² + 23t - 240 = 0
t = 3.213 s
s(t) = (g/2)t² + v₀t
where s=240ft is the space, g is gravity acceleration g ≈ 32.174 ft/s²
v₀=23 ft/s is the initial velocity
Plug data into the formula
s(t) = 16.087t² + 23t
v(t) = ds/dt = 32.174t + 23
after 2 seconds
s = 110.348ft
v = 87.348ft/s
hits the floor when s = 240ft
16.087t² + 23t - 240 = 0
t = 3.213 s
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the position equation is -16t^2+(Initial velocity)t plus height(c)=position
-16t^2+23t+240= position f(t)
-32t+23=velocity f'(t)
-16*4= 23*2+240=
you can figure velocity and set the position equation to 0 to find the time
-16t^2+23t+240= position f(t)
-32t+23=velocity f'(t)
-16*4= 23*2+240=
you can figure velocity and set the position equation to 0 to find the time