Answer:
A) v(t) = (32 - 1.6t) m/s
a(t) = -1.6 m/s²
B) t = 20 seconds
C) 320 m
D) t = 5.85 seconds going up and 34.14 seconds going down
E) 40 seconds
Explanation:
Height is given by the equation;
S(t) = 32t - 0.8t²
A) Velocity after time t is gotten from first derivative of the distance.
Thus;
v(t) = dS/dt = 32 - 1.6t
Acceleration at time t is gotten from derivative of the velocity.
Thus;
a(t) = d²S/dt² = -1.6 m/s²
B) At highest point, velocity is zero.
Thus;
32 - 1.6t = 0
1.6t = 32
t = 32/1.6
t = 20 seconds
C) To find how high the rock goes, it means we are looking for maximum height.
This will be at t = 20 seconds.
Thus;
S(20) = 32(20) - 0.8(20)²
S(20) = 640 - 320
S(20) = 320 m
D) we want to find the time it will take to reach half its maximum height.
since maximum height is 320 m, then half the maximum height is; S_½ = 320/2 = 160
Thus;
160 = 32t - 0.8t²
0.8t² - 32t + 160 = 0
Using quadratic formula, we will get;
t = 5.85 seconds going up and 34.14 seconds going down
E) time the rock is aloft = twice the time it took to reach maximum height.
Thus; t_aloft = 2 × 20 = 40 seconds
