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Hi, here is a sum on projectile motion that I had difficulty understanding. I solved it and for the first part I get a result 2.45sec for t?
which I don't know if correct. And I did not understand b and c. Will anyone be kind enough to explain and solve it for me. Below is a diagram I used as a reference. Thank you
Q) A block of mass m = 3 kg is sliding along a frictionless inclined surface that makes an angle of φ=30 with respect to the horizontal surface. At the lowest points of the inclined surface, a projectile is fired at a speed of v0 = 12 m/s that makes an angle θ=45 with respect to the horizontal. The aim of this problem is to find the time when the projectile will hit the block.
(a) Find the time when the projectile hits the block. (Given, the initial distance between the block and the projectile along the inclined surface at t = 0 is R0 = 10 m) (b) Find the height h, where the projectile will hit the block.
(c) Now assuming that the projectile hits the block at its maximum range R along the inclined surface, Rmax=?
1 Answer
- az_lenderLv 74 weeks ago
To see that your answer for part(a) cannot make any sense at all, consider the altitude of the projectile after 2.45 seconds. If the projectile were shot straight up at 12 m/s, the time for it to fall all the way back to the ground would be (24 m/s)/(9.8 m/s^2) = 2.45 seconds. However, there is no reason to suppose that the projectile was shot straight upward. Indeed we are TOLD that the projectile is shot upward at a 45-degree angle.
The horizontal component of the projectile's speed will be a constant
(12 m/s)*cos(45) = 8.485 m/s,
and the vertical component of the projectile's speed will be
8.485 m/s - (9.8 m/s^2) t.
We seek a time when the projectile's horizontal displacement will be exactly sqrt(3) times its vertical displacement (because that's tan(30 deg)), so we can write
8.485 t = sqrt(3) * [8.485 t - 4.9 t^2] =>
8.485 = sqrt(3) * [8.485 - 4.9 t] =>
8.485*sqrt(3) - 8.485 = 4.9 t =>
t = 8.485*0.732/4.9 = 1.27 seconds.
(b) At this time, the height of the projectile is
(8.485 m/s)*(1.27 s) - (4.9 m/s^2)(1.27 s)^2 = use calculator
(c) I guess they're looking for the horizontal displacement of the projectile at this same time, and that would be (8.485 m/s)*(1.27 s) = use calculator.
