Help me with this math question please!?

Question

If Sam launches a rocket at a 40 degree angle of elevation and it lands 20 meters from the launch pad, how high did the rocket soar?

Answer 1

rocket is under influence of gravity so this is physics question

let v be initial velocity of rocket

let t be its airtime

write 2 simultaneous equations in terms of v and t

(v cos 40) * t = 20 => means that its horizontal velocity mulitplied by the time equals its horizontal displacement (remember a rocket always has constant horizontal veloctiy)  (based on formula d =vt )

0.5t = (v sin 40) / 9.8 => means that time it takes to fall which is half of airtime is equal to its final velocity divided by the acceleration of gravity (based on formula a=v/t so t= v/a)

now solving for the equations you get (substitution is best method, replace the t of second equation into first equation)

v = 14.107573 m/s   t = 1.850647577 s

now time it takes to fall is 0.5t or 0.925323788 seconds

now its max height is found by equation h=0.5at^2

h=0.5(9.8)*0.925323788^2

h= 4.195498152 m

so the rocket soared 4.2 metres high

hope this helps

Answer 2

Distance traveled in horizontal direction, Sh = 20 m

Now from the equations of motion, we have

S = ut + 1/2 at²

Here for motion in horizontal direction,

Intial velocity in horizontal direction = Uh

Acceleration in horizontal direction, a = 0

Time taken for traveling 20m, t = Total time of flight, T

So,

20 = UhT + 0

-> Uh = 20/T

Now, Uh = U cosØ

And we know the angle of elevation Ø = 40°

Therefore, Uh = U cos40

& U = Uh/cos40

Now, From Equations of motion, v = u + at

For vertical motion,

Intial velocity in verical direction, Uv = UsinØ = Usin40

Substituting for U,

Uv = (Uh/cos40)sin40

     = Uh (sin40/cos40)

     = Uh tan40

     = (20/T) tan40

Acceleration in vertical direction = -g (g acting downwards)

Now, we know that at time t=T/2 (half the time of flight) the maximum height is attained & at this height, velocity in vertical direction is zero.

Thus,

Vv = 0 = Uv - gT/2

          = (20/T) tan40 - gT/2

-> gT/2 = 20 tan40/T

-> T² = 40 tan40/g

-> T  = 1.85s

Total time of flight, T = 1.85 s

Therefore, time at which the rocket is at maximum height = T/2

                                                                                    = 0.93s

Again, from equations of motion,

S = ut + 1/2 at²

For vertical motion at time t = T/2, distance traveled S = Maximum height H

H = Uv (T/2) + 1/2 (-g) (T/2)²

   = (20/T tan40)(T/2) - g/2(T/2)²

   = 10 tan40 - g/8 (T)²

   = 4.2 meters (approx.)

Answer 3

you can make a triangle out of this,

your x is 20meters (one side of the triangle), your angle is 40degrees and your y (the height, other side of the triangle) is unknown.

the third side which is the hypotenuse is not given, so we have to use tangent ^_^

So, tan40 = y/x

tan40 = y/20

y = 20 * tan40

y = 16.78meters  (answer)

Hope this helps you! Goodluck ^_^V