- #1
scorpion990
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Galileo throws a rock from the top of the Leaning Tower of Pisa at an upward angle of 60 degrees with speed v. The rock is in flight for 6.5 seconds and hits the ground 15 m from the base of the building. Ignore air resistance and ignore the fact that the tower tilts a bit. a. What is the speed v? b. How high off the ground is the top of the tower?
This is what I did the first time:
R = v^2*sin(theta)/g (Formula for range)
15 = v^2*sin(60)/9.8
v is approximately 13.028468 m/s
vy = 13.028468sin60 = 11.2829843 m/s
I then used the fact that ay = -9.8 m/s^2 and integration to make a model for position. I set it equal to 0, and I got that the height of the building equals 133 m. However, the real height is about 50 m, so I know I'm doing something incorrectly =/
I just realized that my logic is flawed. The "range" is defined as the distance in the x direction traveled where the starting and ending heights are the same. Mine are not, so I can't use the range formula. However, I can't imagine how else I can solve this. Can anybody point me in the right direction? Thanks.
This is what I did the first time:
R = v^2*sin(theta)/g (Formula for range)
15 = v^2*sin(60)/9.8
v is approximately 13.028468 m/s
vy = 13.028468sin60 = 11.2829843 m/s
I then used the fact that ay = -9.8 m/s^2 and integration to make a model for position. I set it equal to 0, and I got that the height of the building equals 133 m. However, the real height is about 50 m, so I know I'm doing something incorrectly =/
I just realized that my logic is flawed. The "range" is defined as the distance in the x direction traveled where the starting and ending heights are the same. Mine are not, so I can't use the range formula. However, I can't imagine how else I can solve this. Can anybody point me in the right direction? Thanks.