Could anyone help me understand/solve these physics problems?

Question

I've tried everything I can think of both for equations with and without constant acceleration/velocity but I feel like I'm not given enough information. Can anyone help me figure out how to solve these problems?

A rock is tossed straight up with a velocity of +20m/s When it returns, it falls into a hole 10m deep. What is the rock's velocity as it hits the bottom of the hole, and how long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

A toy train is pushed forward and released at x0 = 2.0m with a speed of 2.0m/s. It rolls at a steady speed for 2.0s, then one wheel begins to stick. The train comes to a stop 6.0m from the point at which it was released. What is the magnitude of the train's acceleration after it begins to stick? (This one requires 2 significant figures in the answer).

Please don't give an answer without an explanation. Thanks so much!

JullyWum · Accepted Answer

Neglecting air resistance, the rock will have a downward velocity of 20.0m/s as it passes the the throwing point .. u = 20.0 m/s

Apply .. v² = u² + 2as .. (a = accel.due to gravity, 9.80m/s² .. s = hole distance, 10.0m)
v² = 20.0² + (2 x 9.8 x10)
v² = 596 .. .. v = √596 .. .. ►v = 24.40 m/s

Time taken (by a method that avoids a quadratic solution)
Apply [ v = u + at ] to the journey back to throwing point .. (upward direction of initial vel. is +ve)
u = 20.0 m/s .. v = - 20.0 m/s .. a = - 9.80 m/s²
-20.0 = 20.0 - (9.80 x t) .. t = 40 / 9.80 .. .. .. t = 4.08 s

Add time to travel the additional distance  (10.0m) .. (u = 20.0m/s, v = 24.40m/s)
v = u + at'
24.40 = 20.0 + (9.80 x t') .. .. t' = 4.40 /9.80 .. .. t' = 0.45s

Total time = (4.08 + 0.45)s .. .. ►T = 4.53 s

For the train, during it's deceleration period ..
u = 2.0 m/s .. v = 0 .. d = 6.0m -(2m/s x 2s) = 2.0m

v² = u² + 2ad
0 = 2.0² + (2 x a x 2.0) .. .. 0 = 4 + 4a .. .. ►a = - 1.0 m/s²

boice · Answer

D = 55ft. t = sqrt(2d/g) t = sqrt(110ft/32ft/s^2) t = a million.86s I particularly don't know how you obtain 0.33s. Which in case you artwork it out backwards and also you remedy g. g = 2d/t^2 g = 1010ft/s^2 Acceleration = speed / Time speed = Distance / Time surely Acceleration = (Distance / Time) / Time = Distance / (Time)^2 aka Distance over seconds squared. So no, you won't be able to forget with reference to the seconds squared at the same time as fixing for time that's squared. so that you may sq. root the time with the intention to get the truly value of time, and not in any respect time squared.

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Could anyone help me understand/solve these physics problems?

2 Answers