does anyone know how to do this physics?

Question

a basketball is thrown straight upward with an initial velocity of 3 meters per second. (a) by setting the motion equation equal to zero, find how long it takes for the basketball to stop moving upward. (b) use this time in the position equation to find the maximum height which the basketball reaches.

Answer 1

(a)
here,

the initial velocity, u = 3 m/s
the final velocity, v = 0 m/s
gravitational accelaration, g = 9.8 m/s^2

from the motion equation we have,
v = u - gt
or, 0 = 3 - 9.8t
so, t = 3/9.8

t = 0.306 second

(b)

from the postion equation,

s = ut - 1/2 gt^2

= 3 * 0.306 - 1/2 * 9.8 * (0.306)^2

= 0.459 meter

so the maximum height the basketball reaches = 0.459 m

Hope this was helpful.

Answer 2

This is a very simple case where we can use the well known equations of the motion.
First,
You have to use the first equation of motion.
Vf= Vi + gt.
and solve it for the value of "t" using Vi as 3 m/sec. use g as -9.81 meter per second square as the ball is thrown upward.
It will yield as 0.305 seconds
Then put this value of t in the second equation of the motion and solve it for the max. height .i.e "h"
h= Vi t + 1/2 gt^2.
This equation will yield as 0.459 meters, use g = -9.81 m/sec again
I hope this will help,
May Allah bless you.

Answer 3

dont know about the motion equation, but can use vf=vi+at
0=3+(-9.8)t
t=.306seconds
dont know about the position equation either, but can use vf^2=vi^2+2as
0^2=3^2+2(-9.8)s
s=.459meters