- #1
caels
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I missed a few days of class and have a quiz tomorrow so I'm not entirely sure where to start with this. The question is:
A golfer hits a ball (m = 38 g) from a cliff top (h = 46 m) and times how long it takes to splash in the ocean (t = 7.0 s). The ball hits the water at a distance (x = 154 m) from the vertical cliff bottom. Find the ball's initial velocity and the angle of launch.
I have y = 46 m, x = 154 m, t = 7.0 s
I assume I have to use the two-dimensional kinematic equations, but I just can't seem to get anywhere towards finding the initial velocity.
My first thought was
x = v_0x * t
v_0x = 154 / 7 = 22 m/s
Then take
y = (v_0y)t+1/2 gt^2 = 7(v_0y) - 240.34
v_0y = 40.905
And then square both of those and take the square root. But the answer I get doesn't agree with the instructors, so I'm lost.
A golfer hits a ball (m = 38 g) from a cliff top (h = 46 m) and times how long it takes to splash in the ocean (t = 7.0 s). The ball hits the water at a distance (x = 154 m) from the vertical cliff bottom. Find the ball's initial velocity and the angle of launch.
I have y = 46 m, x = 154 m, t = 7.0 s
I assume I have to use the two-dimensional kinematic equations, but I just can't seem to get anywhere towards finding the initial velocity.
My first thought was
x = v_0x * t
v_0x = 154 / 7 = 22 m/s
Then take
y = (v_0y)t+1/2 gt^2 = 7(v_0y) - 240.34
v_0y = 40.905
And then square both of those and take the square root. But the answer I get doesn't agree with the instructors, so I'm lost.