Calculus; Displacement, Velocity, Acceleration problem.?

Question

A coin is dropped from the top of a building 150m high. using a = -9.8 m/s^2.

a. How long will it take for the coin to strike the ground?
b. With what velocity will the coin strike the ground?

Answer 1

This should be pretty simple. I tried it myself. But I got irrational numbers as answers. If that's OK with you, follow this answer.

NOTE: 150/4.9, its square root and square root of 2940 are all listed as approximate values.

u = 0
h = 150
g = 9.8 m/s^2 (You must take it to be positive as acceleration is in the direction of motion, that is downward)

h = ut + (gt^2)/2
h = 0 + 4.9t^2
4.9t^2 = 150
t^2 = 150/4.9
t^2 = 30.58
t = 5.53

Time taken to strike ground is approximately 5.53 seconds

v^2 - u^2 = 2gh
v^2 = 19.6*150
v^2 = 2940
v = 54.22

The coin hits the ground at approximately 54.22 m/s

Answer 2

The velocity is supposed to be negative, since the canister is moving downwards. Velocity and speed are not, as most people seem to believe, the same thing. Speed is the magnitude of velocity (which would be the absolute value of the velocity in this case). Velocity can be negative, but speed cannot. So use the absolute value of the velocity you found and compare it to the maximum impact SPEED.

Answer 3

(a)
y = 0
y(0) = 150

v(0) = 0 (because the coin is DROPPED and not thrown)
a = -9.8 m/s^2

y = y(0) + v(0)*t - [9.8/2]*t^2
0 = 150 - 4.9*t^2
t = 5.53 sec

(b)
v(t)^2 = v(0)^2 - 2*a*h
v(5.53) = sqrt[ 0 - 2*-9.8*150 ]
v(5.53) = 54.22 m/sec