Asked by vien
a projectile is fired at an angle of 27 above the horizontal from the top of a cliff 500 ft high. The initial speed of projectile is 1670 ft/s. How far will it move horizontally before hitting the level ground at the base of the cliff?
Answers
Answered by
Damon
g = about 32 feet/second^2 as I remember from my childhood
so (1/2) g = 16
vertical problem
Vi = 1670 sin 27
Hi = 500
h = Hi + Vi t - 16 t^2
0 = 500 + (1670 sin 27) t - 16 t^2
solve quadratic for t, use the positive answer, that is time in the air
horizontal problem
u = 1670 cos 27 forever
range = u t
so (1/2) g = 16
vertical problem
Vi = 1670 sin 27
Hi = 500
h = Hi + Vi t - 16 t^2
0 = 500 + (1670 sin 27) t - 16 t^2
solve quadratic for t, use the positive answer, that is time in the air
horizontal problem
u = 1670 cos 27 forever
range = u t
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