Given: Initial velocity [tex]v_i= 15.0m/s[/tex]
Final Velocity [tex]v_f=8.0 m/s[/tex]
To find : Height when jumpled 8.0 m/s upwards.
Solution: We already have values of initial velocity and final velocity.
We know, accereration due to gravity is given by [tex]a= -9.8 m/s^2[/tex].
It's negative because when jump it's in opposite direction.
[tex]We know formula, v_f ^2 = v_i ^2+2ah.[/tex]
Where h is the height when jumpled 8.0 m/s upwards.
Plugging values of [tex]v_i, \ v_f \ and \ a \ in \ formula \ above.[/tex]
[tex](8.0)^2 = (15.0)^2 +2(-9.8)*h[/tex]
64= 225 -19.6h
Subtracting both sides by 225.
64-225= 225 -19.6h-225.
We get,
-161 = -19.6h
Dividing both sides by -19.6, we get
[tex]\frac{-161 }{-19.6} =\frac{ -19.6h}{ -19.6}[/tex]
h= 8.2143
Rounding to nearest tenth, we get
h= 8.2 meter.
His height is 8.2 meter when he is jumping 8.0 m/s upwards.
