Question

Please help me complete my home assignment! ​

Answer 1

Answer:

Rocket's maximum height=3626.25 feet.

The Rocket will reach its height after 4 seconds.

After fired the rocket reaches the ground at 30 seconds(more like that only).

Answer 2

Answer:

See the solution with refer to the above image!

☆Points to note:

• at the highest point,the rocket will be momentarily at rest(v=0)

•we can see here that the the displacement eq. with respect to time,can be described from A,

Bcz at height 'A' it has some disp. which will be not vary with time = 800ft

• varying part = -16t²+80t

☆Solution:

1) what is the rocket's max height?

•h(t) = -16t² + 80t + 800

•$\dfrac{d(h)}{dt}$ = -16(2t) + 80(1) + 0

•v = -32t +80

•0= -32t + 80

•t=80/32 = 2.5s (this is time from A to reach B)

》h = -16(2.5)² + 80(2.5) + 800

》$\ h_{max}$ = -100+200+800

》$\bf{ h_{max}}$ = 900ft.

2) time taken to reach max height?

》its in the above part.

》2.5sec.

3)time taken by rocket to reach ground?

》total time = $\ t_{AB}+t_{BD}$

》$\ t_{AB}=2.5s$

》we know that disp. from B to D is 900ft.

• s = ut + 1/2 at²

• -900 = 0 + 1/2(-g)t² (g=32ft/s²)

• 900 = 16 t²

• $\ t = \sqrt{ \dfrac{900}{16}}$

•t= 30/4 = 7.5s

》t = $\ t_{AB}+t_{BD}$

• t= 2.5 + 7.5

• t = 10s

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☆only thank the answer if all the parts are correct,

then i'll know if there is some mistake!!