A golf ball is hit and makes a 45 $\circ$ angle with the horizontal as it leaves the tee.It lands 200 m away.What was the ball's initial speed?
A)35.11 m/s
B)26.82 m/s
C)44.27 m/s
D)56.82 m/s
E)31.44 m/s

Question

Quizplus · Accepted Answer

We can use the kinematic equations to solve for the initial speed, given the launch angle and distance traveled.

The horizontal range, R, is given by:

R = v0^2 * sin(2θ) / g

where v0 is the initial speed, θ is the launch angle, and g is the acceleration due to gravity.

Plugging in the given values, we have:

200 m = v0^2 * sin(2 * 45°) / 9.81 m/s^2

Simplifying:

200 m * 9.81 m/s^2 = v0^2 * sin(90°)

v0 = sqrt(200 m * 9.81 m/s^2 / sin(90°))

v0 = 44.27 m/s

Therefore, the initial speed of the golf ball was 44.27 m/s.

A Golf Ball Is Hit and Makes a 45 ∘\circ∘ Angle with the Horizontal as It Leaves the Tee

A Golf Ball Is Hit and Makes a 45^$\circ$ Angle with the Horizontal as It Leaves the Tee