Question

How long does it take the arrow to come back to the ground if Andrew is an avid archer and he launches an arrow that takes a parabolic path, modeled by the equation $ y = - 4.9{t^2} + 48t $ ?

Answer 1

Hint: To solve this problem we have to observe the motion in projectile motion. From there we have to find the vertical velocity of the arrow and calculate the time of flight using the equation of path given.
In projectile motion a particle launched with initial velocity $ u $ making an angle $ \theta $ with the horizontal, total time of flight of the particle in projectile motion is $ T = \dfrac{{2u\sin \theta }}{g} $ . Velocity of a particle is given by, $ v = \dfrac{{dr}}{{dt}} $ where $ r $ is the position of the particle.

Complete step by step answer:
We have given here the path modeled by the equation $ y = - 4.9{t^2} + 48t $ . Now, it is the path of motion along the Y-axis of the particle. So, for a projectile motion we just have to consider the vertical motion of the projectile and calculate the time of flight of the particle.
Now we know, in projectile motion a particle launched with initial velocity $ u $ making an angle $ \theta $ with the horizontal the maximum height reached by the projectile is $ H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} $ and total time of flight of the particle in projectile motion is $ T = \dfrac{{2u\sin \theta }}{g} $ .
So, if we look at the equation we know that the initial velocity along the vertical is $ u\sin \theta $ . Hence, the total time of flight is $ T = \dfrac{{2{u_y}}}{g} $
Now, we have here the path of the particle along the Y-axis. So, if we differentiate the equation at $ t = 0 $ we will get the initial velocity along the Y-axis or vertical .
So, differentiating the equation of path at $ t = 0 $ we get,
 $ {u_y} = \dfrac{d}{{dt}}{\left. {( - 4.9{t^2} + 48t)} \right|_{t = 0}} $
Or, $ {u_y} = {\left. {( - 2 \times 4.9t + 48)} \right|_{t = 0}} $
up on simplifying we have,
 $ {u_y} = 48 $
So, the initial velocity along vertical is $ {u_y} = 48 $
Now, putting this value in $ T = \dfrac{{2{u_y}}}{g} $ we have total time of flight of the arrow as,
 $ T = \dfrac{{2 \times 48}}{{9.8}} $
Or, $ T = 9.795 $
So, the arrow will come back to the ground in $ 9.795s $ .

Note:
We can also solve this problem using the equation for 1D-motion. Then we will use the formula, $ h = ut - \dfrac{1}{2}g{t^2} $ . Putting $ h = 0 $ and finding the initial velocity as done in the solve previously will yield the time taken by the arrow to touch the ground.