A Physics Question?

Question

An object is propelled vertically upward with an initial velocity of 20 meters per second. The distance s (in meters) of the object from the ground after t seconds is s = -4.9t^2 + 20t.

(1) When will the object reach a height of 100 meters?

(2) What is the maximum height of the object?

Answer 1

1)
-4.9t^2 + 20t = 100
4.9t^2 - 20t + 100 = 0
t = (20 +/- sqrt(400 - 1960)) / 9.8
t has no real value because sqrt(400 - 1960) = sqrt(-1560) is an imaginary number. Therefore, the object does not reach a height of 100 m.

2) We find the extreme value by finding where the derivative is equal to zero.

ds/dt = -9.8t + 20 = 0 ==> t = 20/9.8 = 2.04
s(2.04) = -4.9*2.04^2 + 20*2.04 = 20.4 m

Answer 2

1- Never
2- The max height means s=0 m which means 4.9t^2=20t ==> t=20/4,9=4,0816 second
The max height of the object is integral of the s between 0->4,0816 which is about 55,5324 m.

Answer 3

v= u -gt
for velocity to be zero,
0 = 20 - 9.8 t
9.8 t = 20
t= 20/9.8= 2.04 seconds

max height
s= ut- .5 g t^2
= 20 * 2.04 - 4.9 (2.04)^2
= 40.8 - 20.39
= 20.41 meters

Hence will never reach height of 100 meters

Answer 4

a. vo=20 m/sec g=~10 m/sec^2
at the top v=0 ,,0=vo-g t,,,g t =20,,t=20/10=2 sec.

.max.height=g t^2 /2=10 (2)^2 /2=5x4=20 meters.
(1) smaller than 100 m,,no
(2) h=20 m.