MATH SOLVE

5 months ago

Q:
# I’ll put you as brainliest A model rocket is launched from the ground with an initial velocity of100feet per second. How many seconds later does the rocket hit the ground?

Accepted Solution

A:

The law of an object moving with constant acceleration is[tex]s(t)=s_0+v_0t+\dfrac{1}{2}at^2[/tex]Where [tex]s[/tex] is space, [tex]t[/tex] is time, [tex]s_0[/tex] is the initial position, [tex]v_0[/tex] is the initial velocity and [tex]a[/tex] is the acceleration.In this case, if we choose a reference grid with the vertical axis pointing upwards, the acceleration of gravity will point downwards (and thus be negative). The initial position is zero, because the rocket is on the ground, and the initial velocity is 100 (positive because pointing upwards).So, its law is[tex]h(t)=100t-\dfrac{1}{2}gt^2[/tex](I changed [tex]s[/tex] for [tex]h[/tex] since the rocket is moving vertically, so its position is actually its height. Also, g is the acceleration due to gravity).The rocket hits the ground if its height is zero, so if we set [tex]h=0[/tex] we have[tex]0=100t-\dfrac{1}{2}gt^2 \iff t(100-\dfrac{1}{2}gt)=0[/tex]Solving for t, we have either t=0, or[tex]100-\dfrac{1}{2}gt=0 \iff 100=\dfrac{1}{2}gt \iff 200=gt \iff t=\dfrac{200}{g}[/tex]The solution t=0 means that at the beginning the rocket is on the ground. So, we're interested in the other solution. Considering that g is about 32.2 feet/s^2, we have[tex]t=\dfrac{200}{g}\approx \dfrac{200}{32.2}\approx 6.21[/tex]