1. The problem statement, all variables and given/known data A golfer hits a shot to a green that is elevated 4.0 m above the point where the ball is struck. The ball leaves the club at a speed of 15.0 m/s at an angle of 40.0° above the horizontal, which is the +x axis. It rises in the +y direction to its maximum height and then falls down to the green. Ignore air resistance. What is the algebraic expression for the y component vy of the ball's velocity just before landing on the green? Calculate for the y component vy using this equation. known: v(initial)=15.0m/s a_x= 0 a_y= 9.8 m/s^2 v_x (initial) = cos40*15 = +11.5 m/s 2. Relevant equations i thought it would be v^2=v(initial)^2 + 2ay rearranged to solve for just v(y direction). therefore: square root of v(initial)^2 + 2ay. y being the distance in the vertical direction. also: for v(initial) i took the sin40*15 to get 9.6 m/s. 3. The attempt at a solution when i plug this into my homework website, it tells me i am wrong. but this is the only kinematic that i can think of using only v(initial), y displacement, and acceleration. what's wrong with my equation?
What value did you get? Did you use [itex]v_y^2=(v_0)^2_y+2ay[/itex] or [itex]v_y^2=(v_0)^2_y-2ay[/itex] ?
i used the equation LaTeX Code: v_y^2=(v_0)^2_y-2ay and rearranged it so it would be sqtroot(v(initial)^2 + 2ay). i think i got like -8.9 m/s, but the website told me i was wrong.
v=final u=inital if you use v^{2}=u^{2}+2ay then you are saying that acceleration is acting upwards. The only acceleration in this question is gravity which acts downwards.So, you'd need to use v^{2}=u^{2}-2ay