Question

3. A model rocket is launched straight upward at 58.8 m/s.

Answer 1

Answer:

a). 6 seconds

b). 12 seconds

c). 176.4 meters

Explanation:

a). Equation to be applied to calculate the time taken by the rocket to reach at the peak height,

   v = u - gt

where v = final velocity

u = initial velocity = 58.8 m per sec

g = gravitational pull = 9.8 m per sec²

t = duration of the flight

At the peak height,

v = 0

Therefore, 0 = 58.8 - (9.8)(t)

t =  $\frac{58.8}{9.8}$

 = 6 seconds

b). Total time of flight = 2(Time taken to go up)

                                    = 2×6

                                    = 12 sec

c). Formula to get the peak height is,

   $h=ut-\frac{1}{2}gt^2$

   h = (58.8)6 - $\frac{1}{2}(9.8)(6)^2$

      = 352.8 - 176.4

      = 176.4 meters