Question

a ball traveling at 63 m/s begins to roll up an inclined plane before coming to rest. the ball comes to a stop 7.00 m up the incline. what was the velocity of the ball 7.0 s after starting up the incline? select the appropriate formula: $vec{a}=\frac{deltavec{v}}{delta t}$ $vec{d}=vec{v}_i t+\frac{1}{2}vec{a}t^2$ $vec{d}=vec{v}_f t - \frac{1}{2}vec{a}t^2$ $vec{d}=(\frac{vec{v}_f+vec{v}_i}{2}t)$ $v_f^2 = v_i^2+2ad$ final velocity= (before significant digits) final velocity= m/s select an answer provide your answer using the proper significant digits.

Explanation:

Step1: Find the acceleration

We know the initial velocity $v_i = 63$ m/s, the final velocity $v_f=0$ m/s and the displacement $d = 7.00$ m. Use the formula $v_f^{2}=v_i^{2}+2ad$.
\[0 = 63^{2}+2a\times7\]
\[0 = 3969+14a\]
\[14a=- 3969\]
\[a=\frac{-3969}{14}\approx - 283.5\] m/s²

Step2: Find the velocity at $t = 7.0$ s

Use the formula $v = v_i+at$. Here, $v_i = 63$ m/s, $a=-283.5$ m/s² and $t = 7.0$ s.
\[v=63+( - 283.5)\times7\]
\[v=63 - 1984.5\]
\[v=-1921.5\] m/s

Answer:

-1921.5 m/s