Question

A rocket is launched from a height of 3 m with an initial velocity of 15m/s.When will the rocket hit the ground?The rocket will hit the ground ataapproximately 5.67 s.The rocket will hit the ground atbapproximately 4.15 s.The rocket will hit the ground atсapproximately 3.50 s.The rocket will hit the ground atdapproximately 3.25 s.Check itskip

Answer 1

In order to determine the time the rocket takes to hit the ground, use the following formula for free fall when an object is launches:

`y=y_o+v_ot-(1)/(2)gt^2`

where,

vo: initial velocity = 15m/s

yo: initial height = 3m

g: gravitational acceleration constant = 9.8 m/s^2

y: final height = 0 m (the rocket is on the ground)

Replace the previous values into the equation above and order the equation as a quadratic equation in standard form:

`\begin{gathered} 0=3+15t-(1)/(2)9.8t^2 \\ 0=3+15t-4.9t^2 \\ 4.9t^2-15t-3=0 \end{gathered}`

Next, use the quadratic formula, with a = 4.6, b = 15 and c = -3, to find the solutions for t:

`\begin{gathered} t=\frac{-(-15)\pm\sqrt[]{15^2-4(4.9)(-3)}}{2(4.9)} \\ t=(15\pm16.85)/(9.8) \end{gathered}`

take the positive solution for t (because negative times does not have physical meaning), then:

t = (15 + 16.85)/9.8 = 3.25 s

Hence, the rocket takes approximately 3.25 s to hit the ground