- #1
Mr Davis 97
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Homework Statement
A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club at a speed of 14.0 m/s at an angle of 40.0° above the horizontal . It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the peed of the ball just before it lands.
Homework Equations
##v_x = v_{0x} = \left \| v_0 \right \|\cos\theta##
##v_y^2 = v_{0y}^2 + 2ay##
##v = \sqrt{v_x^2 + v_y^2}##
The Attempt at a Solution
We begin with ##v = \sqrt{v_x^2 + v_y^2}## since the magnitude of the final velocity is what we desire. Thus, we need to find ##v_x## and ##v_y##. Therefore, we use ##v_x = v_{0x} = \left \| v_0 \right \|\cos\theta## since the velocity in the x-direction does not change from the initial (we're neglecting air resistance). Thus ##v_x = (14 ~m/s)\cos40^{\circ} = 10.7~m/s##. Now,we find ##v_y##, the final velocity in the y-direction. Since time is not metioned, we'll use ##v_y^2 = v_{0y}^2 + 2ay##, specifically ##v_y = -\sqrt{v_{0y}^2 + 2ay}##. Thus ##v_y = -\sqrt{v_{0y}^2 + 2ay} = -\sqrt{(v\sin\theta)^2 + 2(-9.8~m/s^2)(3~m - 0~m)} = -4.71~m/s##. Therefore, the magnitude of the final velocity should be ##v = \sqrt{v_x^2 + v_y^2} = \sqrt{(10.7~m/s)^2 + (-4.71~m/s)^2} = 11.7~m/s ##. Is this the correct answer? If not what am I doing wrong and why?