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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. 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 and the block is not sliding at all. What is the maximum range along the inclined surface, Rmax=?
2 Answers
- ?Lv 74 weeks agoFavorite Answer
Hey, were these workings helpful?
- oubaasLv 74 weeks ago
a)
block acceleration a = g*sin 30° = 9.8*0.5 = 4.9 m/sec^2
12*0.707*t-4.9t^2 = sin 30°*10-a/2*t^2*sin 30
-4.9/4t^2+4.9t^2-8.48t+5 = 0
3.675t^2-8.48t+5 = 0
t = (8.48+√8.48^2-20*3.675)/7.35 = ?? quantity under root is negative , therefore there is no solution and the box can't be hitten
c) Rmax ??f
for an angle of 30°, the ratio x/h is equal to 1.73 and the ratio h/x = 0.578
8.484t-4.9t/2 = 0.578*(8.484*t)
8.484t-4.9t^2 = 4.904t
3.58t = 4.9t^2
t cancels
t = 3.58/4.9 = 0.73 sec
x = 8.484*0.73 = 6.20 m
h = 6.20*0.578 = 3.58 m
Rmax = h*2 = 3.58*2 = 7.16 m
The real graph shows how intersection between the projectile and the incline happens while projectile is still ascending !!