Question

A ball is thrown upward with an initial velocity of 25 m/s at an angle 30° with the horizontal. The thrower is at the top of a long hill Ich slopes downward at an angle of 15°. How far down the slope does the ball  strike?

Answer 1

vox = vo*cos30 = 25*cos30 = 21.65 m/s

voy = vo*sin30 = 25*sin30 = 12.5 m/s

initial position = yo = 0

after time t it strikes a point x,y

final position = (x , y) = l*cos15 , l*sin15

along horizantal

x = vox*t

t = x/vox .....(1)

along vertical

ay = -g = -9.8 m/s^2

y = -l*sin15

from equations of motion

y = voy*t + 0.5*ay*t^2

-l*sin15 = voy*x/vox - 0.5*g*x^2/vox^2

-l*sin15 = tan30*l*cos15 - 0.5*g*l^2*(cos15)^2/vox^2

-sin15 = tan30*cos15 - 0.5*9.8*l*(cos15)^2/21.65^2

-0.26 = 0.56 - 0.0097*l

l = 84.54 m   <<<---------------answer