QuestionJuly 23, 2025

A rock thrown vertically upward from the surface of the moon at a velocity of 16m/sec reaches a height of s=16t-0.8t^2 meters 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. Find the rock's velocity at time t. v=16-1.6tm/s Find the rock's acceleration at time t. a=square m/s^2 (Simplify your answer. Use integers or decimals for any numbers in the expression.)

A rock thrown vertically upward from the surface of the moon at a velocity of 16m/sec reaches a height of s=16t-0.8t^2 meters 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. Find the rock's velocity at time t. v=16-1.6tm/s Find the rock's acceleration at time t. a=square m/s^2 (Simplify your answer. Use integers or decimals for any numbers in the expression.)
A rock thrown vertically upward from the surface of the moon at a velocity of 16m/sec reaches a height of s=16t-0.8t^2 meters 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. Find the rock's velocity at time t. v=16-1.6tm/s Find the rock's acceleration at time t. a=square m/s^2 (Simplify your answer. Use integers or decimals for any numbers in the expression.)

Solution
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Answer

a. a = -1.6 \, m/s^{2} ### b. 10 sec ### c. 80 meters ### d. 5 sec ### e. 20 sec Explanation 1. Differentiate to find velocity Velocity v(t) is the derivative of height s(t). v(t) = \frac{d}{dt}(16t - 0.8t^2) = 16 - 1.6t. 2. Differentiate to find acceleration Acceleration a(t) is the derivative of velocity v(t). a(t) = \frac{d}{dt}(16 - 1.6t) = -1.6. 3. Find time to reach highest point Set velocity v(t) = 0: 16 - 1.6t = 0 \Rightarrow t = 10 sec. 4. Calculate maximum height Substitute t = 10 into s(t): s(10) = 16(10) - 0.8(10)^2 = 80 meters. 5. Time to reach half maximum height Half maximum height is 40 meters. Solve 16t - 0.8t^2 = 40: 0.8t^2 - 16t + 40 = 0. Use quadratic formula t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} with a = 0.8, b = -16, c = 40. t = 5 sec or t = 10 sec (only t = 5 sec is valid for half height). 6. Total time aloft Solve s(t) = 0: 16t - 0.8t^2 = 0 \Rightarrow t(16 - 0.8t) = 0. Solutions are t = 0 and t = 20 sec.

Explanation

1. Differentiate to find velocity<br /> Velocity $v(t)$ is the derivative of height $s(t)$. $v(t) = \frac{d}{dt}(16t - 0.8t^2) = 16 - 1.6t$.<br /><br />2. Differentiate to find acceleration<br /> Acceleration $a(t)$ is the derivative of velocity $v(t)$. $a(t) = \frac{d}{dt}(16 - 1.6t) = -1.6$.<br /><br />3. Find time to reach highest point<br /> Set velocity $v(t) = 0$: $16 - 1.6t = 0 \Rightarrow t = 10$ sec.<br /><br />4. Calculate maximum height<br /> Substitute $t = 10$ into $s(t)$: $s(10) = 16(10) - 0.8(10)^2 = 80$ meters.<br /><br />5. Time to reach half maximum height<br /> Half maximum height is $40$ meters. Solve $16t - 0.8t^2 = 40$: $0.8t^2 - 16t + 40 = 0$. Use quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 0.8$, $b = -16$, $c = 40$. $t = 5$ sec or $t = 10$ sec (only $t = 5$ sec is valid for half height).<br /><br />6. Total time aloft<br /> Solve $s(t) = 0$: $16t - 0.8t^2 = 0 \Rightarrow t(16 - 0.8t) = 0$. Solutions are $t = 0$ and $t = 20$ sec.
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