Question

A golf player practicing his drives on the third floor of Top Golf hits the ball with a speed of 14m/s a height of 26m above the ground. How far did his ball go?

Answer 1

The launch of projectiles allows to find the result for the horizontal distance traveled is:

•  The distance is:  x = 32.2 m

Projectile launching is an application of kinematics for the motion of bodies where there is no acceleration on the x-axis and on the y-axis we have gravity acceleration.

A reference system is a coordinate system with respect to which measurements are made, in the attachment we have a diagram of the movement.

They indicate that the player is at y₀ = 26m and throws the ball at a velocity of v₀ₓ = 14 m/s, asks the distance traveled when reaching the ground.

When it reaches the ground, its height is zero and since it was thrown horizontally, its initial vertical velocity is zero.

          y = y₀ + $v_{oy}$ t - ½ g t²

          0 = y₀ + 0 - ½ g t²

          t = $\sqrt{\frac{2y_o}{g} }$  

Let's calculate.

         t = $\sqrt{\frac{2 \ 26}{9.8} }$  

         t = 2.30 s

Since time is a scalar it is the same for horizontal displacement.

         x = v₀ₓ t

Let's calculate

         x = 14 2.30

         x = 32.2 m

In conclusion, using the launch of projectiles we can find the result for the horizontal distance traveled is:

• The distance is:  x = 32.2 m.

Learn more here:  brainly.com/question/12792922