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EaGlE
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A man stands on the roof of a building of height 16.3m and throws a rock with a velocity of magnitude 31.6m/s at an angle of 28.1 degrees above the horizontal. You can ignore air resistance
A.) Calculate the maximum height above the roof reached by the rock. Take free fall acceleration to be 9.80 m/s^2.
ok first i found vectors x and y at 28.1 degrees
cos(28)*31.6 = vector x
sin(28)*31.6 = vector y
x = 27.8752
y= 14.883
using this formula:
x(t) = x(0) + v(0)t + 1/2at^2
x(t) = 16.3 + (31.6m/s)t + 1/2(-9.8)t^2 <--- is that right how i set it up? is a a negative or positive number? and will there be two times? i know I am doing something wrong, i just don't know what...if i solve for t now, there will be two times, which i don't know what to do with it.
B.) Calculate the magnitude of the velocity of the rock just before it strikes the ground. Take free fall acceleration to be 9.80 m/s^2
A.) Calculate the maximum height above the roof reached by the rock. Take free fall acceleration to be 9.80 m/s^2.
ok first i found vectors x and y at 28.1 degrees
cos(28)*31.6 = vector x
sin(28)*31.6 = vector y
x = 27.8752
y= 14.883
using this formula:
x(t) = x(0) + v(0)t + 1/2at^2
x(t) = 16.3 + (31.6m/s)t + 1/2(-9.8)t^2 <--- is that right how i set it up? is a a negative or positive number? and will there be two times? i know I am doing something wrong, i just don't know what...if i solve for t now, there will be two times, which i don't know what to do with it.
B.) Calculate the magnitude of the velocity of the rock just before it strikes the ground. Take free fall acceleration to be 9.80 m/s^2
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