Answer:
Explanation:
Given
height of building (h)=60 m
Range of ball=100 m
(a)time travel to cover a vertical distance of 60 m
[tex]h=ut+\frac{at^2}{2}[/tex]
[tex]60=0+\frac{9.8\times t^2}{2}[/tex]
[tex]t^2=12.24[/tex]
t=3.49 s
(b)To cover a range of 100 m
R=ut
[tex]100=v_x\times 3.49[/tex]
[tex]v_x=28.57 m/s[/tex]
(c)vertical component of velocity just before it hits the ground
[tex]v_y=u+at[/tex]
[tex]v_y=0+9.81\times 3.49=34.202 m/s[/tex]
(d)[tex]v_{net}=\sqrt{v_y^2+v_x^2}[/tex]
[tex]v_{net}=\sqrt{1986.02}=44.56 m/s[/tex]
