You want to jump from an 8.2 m high building to a 10 m high building. The gap between the buildings is 2.4 m, and you can run fast enough so that your speed as you leave the edge of
the lower building is 7.0 m/s. What range of launch angles will allow you to reach the top of the
second building?
I don't even know where to start with this. I know that the horizontal velocity would be 7cos(theta) and the vertical velocity would be 7sin(theta)-9.8t. Which then allows me to find that the horizontal position is 7cos(theta)t, and the vertical position is 7sin(theta)t-4.9t^2. What I don't know is how to find if what angle range leaves the horizontal position greater than 2.4 and the vertical position is greater than 1.8. Any help is great, thank you.
2 answers
Lol, do you go to ISU or something?
Range = Vo^2*sin(2A)/g.
2.4 = 7^2*sin(2A)/9.8.
A = 14.7o. = Minimum launching angle.
Max. range occurs at 45o:
Range = 7^2*sin(2*45)/9.8 = 5 m., max.
Range of launch angles = 14.7 t0 45o.
2.4 = 7^2*sin(2A)/9.8.
A = 14.7o. = Minimum launching angle.
Max. range occurs at 45o:
Range = 7^2*sin(2*45)/9.8 = 5 m., max.
Range of launch angles = 14.7 t0 45o.