you would draw the picture and get the initial y velocity.
then, use any one of the possible equations to get the max height it goes. (at max height, velocity in the y direction is 0)
when you get that height, subtract it from the max to get the initial platform height
A toy rocket is fired off a platform with an initial speed of 223 ft/sec at an angle of elevation of 32 degrees and achieves a maximum height of 291 ft. How how above the ground was the platform?
3 answers
Can you give us the weight of the toy rocket
The distance Y that is rises above the platform is given by
M g Y = (1/2)M (Vsin32)^2
Y = [1/(2g)](118.2 ft/s)^2 = 217 ft
The above is a conservation of energy equation that uses only the vertical-motion part of the kinetic energy. The Vcos32 part of the KE does not change.
Since the toy rocket rises 291 ft from ground level, the platform must be 74 feet high.
I used 32.2 ft/s^2 for g
M g Y = (1/2)M (Vsin32)^2
Y = [1/(2g)](118.2 ft/s)^2 = 217 ft
The above is a conservation of energy equation that uses only the vertical-motion part of the kinetic energy. The Vcos32 part of the KE does not change.
Since the toy rocket rises 291 ft from ground level, the platform must be 74 feet high.
I used 32.2 ft/s^2 for g
2 answers