The projectile's altitude at time [tex]t[/tex] is given by
[tex]y=200\,\mathrm m-\dfrac g2t^2[/tex]
where [tex]g=9.8\,\frac{\mathrm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. It reaches the ground when [tex]y=0[/tex], at which point we find
[tex]0=200\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\implies t=6.4\,\mathrm s[/tex]
so the answer is C.
Answer:
Projectile reaches ground level after 6.4 seconds
Explanation:
Uniformly varied rectilinear motionor also called accelerated uniform rectilinear motion, is motion characterized by having a straight line path and a constant and non-zero acceleration, therefore the speed in this motion changes uniformly depending on the direction of its acceleration.
This is what happens in this case. In this type of moving the equation used to calculate a distance traveled is:
d = Vo * t + [tex]\frac{1}{2}[/tex]*a * t²
where Vo = initial velocity, a = acceleration, d = distance traveled
In this case:
Replacing:
[tex]200m=\frac{1}{2} *9.8\frac{m}{s^{2} } *t^{2}[/tex]
Solving for t you get:
[tex]t=\sqrt{\frac{200}{\frac{1}{2} *9.8} }[/tex]
t= 6.388 seconds≅ 6.4 seconds
Projectile reaches ground level after 6.4 seconds
