Uncovering Projectile Motion Properties

In summary, a projectile with unknown initial velocity lands 19.8 s later on the side of a hill, 2710 m away horizontally and 463 m vertically above its starting point. To determine the initial velocity, we can use the 2D kinematics equations and solve for the vertical and horizontal components. The maximum height above the launch point can be found by finding the time when the vertical component of velocity is 0. As it hits the hill, the projectile has a certain speed and direction, which can be determined from the components of velocity at t = 19.8s.
  • #1
creativeone
6
0
A projectile, fired with unknown initial velocity, lands 19.8 s later on the side of a hill, 2710 m away horizontally and 463 m vertically above its starting point. (Ignore any effects due to air resistance.)

(a) What is the vertical component of its initial velocity?
(b) What is the horizontal component of its initial velocity?
(c) What was its maximum height above its launch point?
(d) As it hit the hill, what speed did it have and what angle did its velocity make with the vertical?




Known 2D kinematics equations



3. How would you do this, I have no idea.
 
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  • #2
It's just like the baseball problem you did.
Start with the y component.
[itex] y = y_0 + v_y t + \frac{1}{2} a t^2 [/itex]

Assuming [itex] y_0 = 0 [/itex]
[itex] y(t=19.8 s) = 463 m = v_y (19.8s) + \frac{1}{2} (-9.8 m/s) (19.8s)^2 [/itex]

Part b is similar.

Part c. Find the time when v_y = 0.

Part d. From the components of velocity at t = 19.8s, what is the speed? What is the direction?
 

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