A shell of mass 2 kg is shot upward with an initial velocity of 100 m/sec. The magnitude of the force due to air resistance is M 20 Suppose that the vertical position axis x is oriented upwa with x = 0 at ground level. Denote by x(t) and v(t) the position and velocity of the shell at time t (in seconds). The acceleration due to gravity has magnitude g=9.81 m/sec². The following numerical values might be useful in the calculations: In 1.96=0.764, In 98.04 =4.58, In 392.4 = 5.97, In 492.4 = 6.20
A shell of mass 2 kg is shot upward with an initial velocity of 100 m/sec. The magnitude of the force due to air resistance is M 20 Suppose that the vertical position axis x is oriented upwa with x = 0 at ground level. Denote by x(t) and v(t) the position and velocity of the shell at time t (in seconds). The acceleration due to gravity has magnitude g=9.81 m/sec². The following numerical values might be useful in the calculations: In 1.96=0.764, In 98.04 =4.58, In 392.4 = 5.97, In 492.4 = 6.20
Chapter3: Polynomial Functions
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
Transcribed Image Text:Suppose that the vertical position axis x is oriented upward
A shell of mass 2 kg is shot upward with an initial velocity of 100 m/sec. The magnitude of the force due to air resistance is
with x = 0 at ground level. Denote by x(t) and v(t) the position and velocity of the shell at time t (in seconds). The acceleration due to gravity has magnitude g=9.81 m/sec².
20
The following numerical values might be useful in the calculations:
In 1.96= 0.764, In 98.04 =4.58, In 392.4 = 5.97, In 492.4 = 6.20
(a) Find the differential equation satisfied by v(t)
O A. dv V
+
dt 40
= -9.81

Transcribed Image Text:(c) The position x(t) of the shell is:
OA
40
x(t) = -392 t + 19696 e - 19696
OB.
+9.804
OC. None of the given answers is correct.
OD.
O E.
x(t) = 1.962 t -9.804 e
x(t) =
- 10 t
3
2
(10000 + 392.4 t)² -100³
588.6
x(t) = -392.4 t - 19696 e
t
40
+ 19696
(d) When will the shell reach its maximum height above the ground?
OA. None of the given answers is correct.
OB. The shell reaches the maximum height when t = 9.08 sec
OC. The shell reaches the maximum height when t= 25.48sec
OD. The shell reaches the maximum height when t= 3.9sec
OE. The shell reaches the maximum height when t = 11.2 sec
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