A golf ball rolls off a horizontal cliff with an initial speed of 14.1 m/s. The ball falls a vertical distance of 14.0 m into a lake below. (a) How much time does the ball spend in the air? (b) What is the speed v of the ball just before it strikes the water?
a. hf=hi-1/2 9.8 t^2 solve for t.
b. initial PE+initial Ke=finalPE+finalKE
solve for vf from that.
t=3.1s
a t=3.1s b h=57.5 c v=45m/s
To find the time the golf ball spends in the air, we can use the equation of motion in the vertical direction. The equation is:
Δy = v₀y * t + (1/2) * a * t²
where:
Δy is the vertical distance the ball falls (14.0 m),
v₀y is the initial vertical velocity (0 m/s since the ball is only rolling off a horizontal cliff),
a is the acceleration due to gravity (-9.8 m/s²), and
t is the time spent in the air (which is what we want to find).
Substituting the known values into the equation, we have:
14.0 m = 0 * t + (1/2) * (-9.8 m/s²) * t²
Rearranging the equation, we get:
4.9 t² = 14.0
Dividing both sides by 4.9, we have:
t² = 2.86
Taking the square root of both sides, we find:
t = √(2.86) ≈ 1.69 s
Therefore, the ball spends approximately 1.69 seconds in the air.
To find the speed of the ball just before it strikes the water, we can use the equation of motion in the horizontal direction. Since there is no horizontal acceleration, the horizontal speed of the ball remains constant. The equation is:
Δx = v₀x * t
where:
Δx is the horizontal distance (which is unknown),
v₀x is the initial horizontal velocity (14.1 m/s, given in the question), and
t is the time spent in the air (1.69 seconds, found in part (a)).
Rearranging the equation, we have:
Δx = v₀x * t
Δx = 14.1 m/s * 1.69 s
Δx ≈ 23.83 m
Therefore, the horizontal distance covered by the ball is approximately 23.83 meters. Since the ball is falling vertically, the horizontal speed does not change, which means the speed of the ball just before it strikes the water is also approximately 14.1 m/s.