Question

A basketball rolls onto the court with a speed of
4
.
0
m
s
4.0
s
m
​ 4, point, 0, space, start fraction, m, divided by, s, end fraction to the right, and slows down with a constant acceleration of
0
.
5
0
m
s
2
0.50
s
2

m
​ 0, point, 50, space, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction over
1
4
m
14m14, space, m.
What is the velocity of the basketball after rolling for
1
4
m
14m14, space, m?

Answer 1

Answer:

5.48 m/s

Explanation:

We can find the final velocity of the basketball by using the SUVAT equation:

$v^2 - u^2 = 2ad$

where

v is the final velocity

u = 4.0 m/s is the initial velocity

$a=0.50 m/s^2$ is the acceleration

d = 14 m is the distance covered

Solving the equation for v, we find:

$v=\sqrt{u^2 +2ad}=\sqrt{4.0^2+2(0.50)(14)}=5.48 m/s$

Answer 2

Answer:

1.4 m/s

Explanation:

It on khan academy, i tried 5.48 the answer below and it was wrong