Question

At time t = 0, a projectile is launched from ground level. At t = 2.00 s, it is displaced d = 54 m horizontally and h = 68 m vertically above the launch point. What are the (a) horizontal and (b)
vertical components of the initial velocity of the projectile? (c) At the instant it reaches its maximum height above ground level, what is its horizontal displacement D from the launch point?

Answer 1

you are given,
t=2s
d=54m --horizontal
h= 68m--vertical
to find the horizontal initial velocity, use the formula \(v_{ix} =\frac{d}{t}\)
to find the vertical initial velocity, use one of the five key equations , \(h=v_{iy} t+\frac{1}{2}at^2\)

for letter c,
you need to find the time when it reaches the maximum height..
once you calculated the initial vertical velocity, you can use this to calculate for the time at maximum height
also note that, when an object reaches the maximum height, the final velocity will be zero
then you can use this
\(v_{fy}=v_{iy}+at\)

after you calculate for the time, use the formula \(v_{ix} =\frac{d}{t}\), to find the range at this point.

Answer 2

so vix=54/2 and h=viyt+(1/2)at^2 would be 68=viy(2)+(1/2)(27)(2)^2? and just solve, am i right?

Answer 3

everything is right, except for the value of your acceleration , where did you get 27?
remember it's a projectile, you should use the acceleration due to gravity

Answer 4

ok, i get it

Answer 5

thanks

Answer 6

no problem :)