Question

A golf ball is at rest on a horizontal surface 1.3m from a hole 

A golfer hits the ball so that it moves horizontally with an initial velocity of 1.8 ms power of -1.The ball experiences a constant deceleration of 1.2 ms power od -2 as it travels to the hole .Calculate the velocity of the ball when it reaches the edge of the table 

Answer 1

Explanations & Calculations

Available data:

1. $\small v_0 =1.8ms^{-1}$
2. $\small a= -1.2ms^{-2}$
3. $\small x=1.3m$
4. $\small v_f = ?$

• since the ball moves under constant acceleration, applying a suitable one of the 4 motion equations would give the result as those are valid for motions with constant accelerations.
• Apply $\small v^2 =v_0^2 +2a(x-x_0)$ in this case.

$\qquad\qquad \begin{aligned} \small v_f^2&=\small (1.8ms^{-1})^2+2(-1.2ms^{-2})(1.3-0)\\ &=\small 0.12 \,m^2s^{-2}\\ \small v_f&=\small \sqrt{0.12\,m^2s^{-2}}\\ \small v_f&=\small \bold{0.35\,ms^{-1}} \end{aligned}$

• Note the reduced velocity due to the deceleration.