# Question #b52a9

Jan 12, 2018

Time of flight=7.03 sec
100 feet above the ground at 4.2 and 5.95 sec
Max height is 196 feet,reached with in 3.5 sec

#### Explanation:

Just analysing the equation,you can arrange it in the format of
S= $u \cdot t$-$\left(\frac{1}{2}\right)$a$\left({t}^{2}\right)$
And it comes out to be,
(S-5)= $112 \cdot t$-$\left(\frac{1}{2}\right)$32*$\left({t}^{2}\right)$
That means,this toy rocket is being pulled down by an acceleration of 32 ft/${\sec}^{2}$ i.e acceleration due to gravity in FPS,and at t=0 it is 5 feet above the ground,which satisfies the given information.

So,let us put S=-5 which will tell us after which time the rocket will come back to the ground and it comes out to be t=7.03 sec.(we put -5 because according to the equation upward displacement has been considered as positive)

Similarly we can put S=105 in the equation which will tell us after what time the rocket will be 100 feet above the ground,and solving we get,t=4.2 sec and 5.95 sec.Two values suggest one during going up and the other while coming down it will be again 100 feet above the ground.

Now easily we can find out the time that it will take to reach the highest point using $v = u - a t$(all symbols are bearing their conventional meaning)
So t is 3.5 sec

Now put this in the given equation,you get S=196 feet(don't forget to subtract 5 as it was already 5 feet above the ground)