A rock thrown vertically upward from the surface of the moon at a velocity of 32m/sec reaches a height of sequals32 t minus 0.8 t squaredmeters in t sec. a. Find the rock's velocity and acceleration at time t. b. How long does it take the rock to reach its highest point?c. How high does the rock go?d. How long does it take the rock to reach half its maximum height?e. How long is the rock aloft?

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

the answer is b, have negligible mass

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