A golf ball is hit with an initial velocity of 135 feet per second at an angle of 22 degrees above the horizontal. Will the ball clear a 25 foot wide sand trap whose nearest edge is 300 feet from the golfer?

Question

Will be through with all the word problems soon. I have managed to actually work some on my own!

Thanks.

Answer 1

To determine if the ball will clear the sand trap, we need to find the height of the ball when it reaches the nearest edge of the sand trap.

First, we can find the time it takes for the ball to reach the nearest edge of the sand trap. We can use the horizontal component of the initial velocity and the horizontal distance to the sand trap:

Vx = V * cos(angle)
Vx = 135 * cos(22°)
Vx ≈ 126.39 ft/s

Horizontal distance = Vx * t
300 = 126.39 * t
t ≈ 2.37 seconds

Next, we can find the vertical component of the initial velocity:

Vy = V * sin(angle)
Vy = 135 * sin(22°)
Vy ≈ 50.71 ft/s

Now we can find the height of the ball at the nearest edge of the sand trap using the vertical component of the velocity:

y = Vy*t - 0.5*g*t^2
y = 50.71*2.37 - 0.5*32*2.37^2
y ≈ 120.14 feet

Since the height of the ball (120.14 feet) is greater than the 25-foot wide sand trap, the ball will clear the sand trap.