Question

A golfer hits a ball at an angle of 12 degrees above horizontal. The velocity of the ball is 37 m/s. What is the horizontal distance the ball travels in 2 seconds? A. 9.6 m B. 36 m C. 72 m D. 74 m

Answer 1

72 m is the horizontal distance the travels.

Option: C

Explanation:

The units to express the horizontal and vertical distances are meters (m). The "horizontal" and "vertical" velocities are expressed in "meters per second" (m/s).

Horizontal distance = (initial horizontal velocity)(time)

We can now get the range x from the horizontal component of velocity

$x=v_{x o} \times t$      equation (1)

$\mathrm{v}_{\mathrm{xo}}=\text { initial horizontal velocity }(\mathrm{m} / \mathrm{s})$

x = horizontal distance (m)

t = time (s)

$\mathrm{v}_{\mathrm{x} \mathrm{o}}=\mathrm{v} \cos \theta$

We know that, V = 37 m/s, θ = 12 degree and t = 2 seconds.

To find, $\mathrm{v}_{\mathrm{x} \mathrm{o}}=\mathrm{v} \cos \theta$

$\mathrm{V}_{\mathrm{x} \mathrm{o}}=37 \times \cos 12$

$\mathrm{V}_{\mathrm{x} \mathrm{o}}=37 \times 0.974$

$\mathrm{V}_{\mathrm{x} \mathrm{o}}=36.038$

$\mathrm{Now}, \mathrm{x}=\mathrm{v}_{\mathrm{xot}}$

$x=36.038 \times 2$

x = 72.076 m ~ 72 m

x = 72 m

The horizontal distance is 72 m.