- #1

gjh

- 10

- 6

- Homework Statement
- A soccer player can kick the ball 28 m on level ground, with initial velocity at 40 degrees to the horizontal. At the same initial speed and angle to the horizontal, what horizontal distance can the player kick the ball on a 15 degree upward slope.

- Relevant Equations
- trajectory equation: y = x*tan (theta)-[(g*x^2)/(2*v()^2*(cos(theta)^2) ... (1)

We know x = R =max range (28m) on level ground. Need to find v()^2. Subbing y=0 into (1) above, get v(0)^2 = (gR^2/)/(2*cos (theta)^2 * tan (theta). ... (2)

This didn't seem right, since this means v(0)^2 is a negative number ... maybe my orientation or algebra wrong?

Anyway, didn't see any other way to proceed, so I subbed value of v(0)^2 into (1) above, and let y = x*tan (theta 1) to represent the slope of the hill. Got substantial cancellation and final expression of x = R*[tan (theta (0) - tan (theta (1)]/tan (theta 0), where theta (0) is 40 degrees and theta (1) = 15 degrees. Then x = 19 m, which agrees with the numerical answer in the back of the book ... BUT don't understand how I can use an invalid interim result [square of a number is negative) to obtain a valid final result. Comments, please!

This didn't seem right, since this means v(0)^2 is a negative number ... maybe my orientation or algebra wrong?

Anyway, didn't see any other way to proceed, so I subbed value of v(0)^2 into (1) above, and let y = x*tan (theta 1) to represent the slope of the hill. Got substantial cancellation and final expression of x = R*[tan (theta (0) - tan (theta (1)]/tan (theta 0), where theta (0) is 40 degrees and theta (1) = 15 degrees. Then x = 19 m, which agrees with the numerical answer in the back of the book ... BUT don't understand how I can use an invalid interim result [square of a number is negative) to obtain a valid final result. Comments, please!