Question

a rock thrown vertically upward from the surface of the moon at a velocity of 40 m/sec reaches a height of s = 40t - 0.8t^2 meters in t sec. a. find the rocks velocity and acceleration at time t. v = 40 - 1.6t m/s a = 2.5 m/s^2 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? (simplify your answer. use integers or decimals for any numbers in the expression.)

Explanation:

Step1: Recall velocity - time formula

The velocity - time formula for vertical motion is $v = v_0+at$, where $v_0$ is the initial velocity, $a$ is the acceleration and $t$ is the time. Given $v_0 = 40$ m/s and $a=- 1.6$ m/s², the velocity at time $t$ is $v=40 - 1.6t$ m/s.

Step2: Find time to reach highest - point

At the highest - point, the velocity $v = 0$. Using the formula $v = v_0+at$, we set $v = 0$, $v_0 = 40$ m/s and $a=-1.6$ m/s². Then $0 = 40-1.6t$. Solving for $t$ gives $t=\frac{40}{1.6}=25$ s.

Step3: Find the height formula

The height formula is $s = v_0t+\frac{1}{2}at^{2}$, with $v_0 = 40$ m/s and $a=-1.6$ m/s², so $s = 40t-0.8t^{2}$.

Step4: Find the maximum height

Substitute $t = 25$ s into the height formula $s = 40t-0.8t^{2}$. Then $s=40\times25-0.8\times25^{2}=1000 - 0.8\times625=1000 - 500 = 500$ m.

Step5: Find time to reach half - maximum height

Set $s = 250$ (half of 500) in the height formula $s = 40t-0.8t^{2}$. So, $0.8t^{2}-40t + 250=0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 0.8$, $b=-40$ and $c = 250$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-40)^{2}-4\times0.8\times250=1600 - 800 = 800$. Then $t=\frac{40\pm\sqrt{800}}{1.6}=\frac{40\pm20\sqrt{2}}{1.6}=\frac{40\pm28.28}{1.6}$. We get two solutions: $t_1=\frac{40 + 28.28}{1.6}\approx42.7$ s and $t_2=\frac{40 - 28.28}{1.6}\approx7.3$ s. The smaller value $t\approx7.3$ s is the time to reach the first - half of the maximum height on the way up.

Step6: Find time aloft

The rock comes back to the starting point when $s = 0$. Set $s = 0$ in $s = 40t-0.8t^{2}$, then $0.8t^{2}-40t=0$. Factor out $t$: $t(0.8t - 40)=0$. We have $t = 0$ (initial time) and $0.8t-40 = 0$, which gives $t = 50$ s.

Answer:

a. $v = 40-1.6t$ m/s
b. $t = 25$ s
c. $s = 500$ m
d. $t\approx7.3$ s
e. $t = 50$ s