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No idea how to start.

  • #1
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.
 

Answers and Replies

  • #2
157
0
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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