Answer:
(A). The time is 5.47 sec.
(B). The speed of the rock just before it strikes the ground is 39.59 m/s.
Explanation:
Given that,
Initial velocity = 14.0 m/s
Height = 70.0 m
(A). We need to calculate the time
Using second equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]70=-14\times t+\dfrac{1}{2}\times9.8\times t^2[/tex]
[tex]4.9t^2-14t-70=0[/tex]
[tex]t =5.47\ sec[/tex]
(B). We need to calculate the speed of the rock just before it strikes the ground
Using third equation of motion
[tex]v^2=u^2+2gs[/tex]
Put the value into the formula
[tex]v^2=(14)^2+2\times9.8\times70[/tex]
[tex]v^2=1568[/tex]
[tex]v=\sqrt{1568}[/tex]
[tex]v=39.59\ m/s[/tex]
Hence, (A). The time is 5.47 sec.
(B). The speed of the rock just before it strikes the ground is 39.59 m/s.