Question

A model rocket is launched from point A with an initial velocity v of 86 m/s. If the rockets descent parachute does not deploy and the rocket lands 104m from A, determine (a)the angle that v forms with the vertical, (b)the maximum height h reached by the rocket, and (c)the duration of the flight.

Answer 1

Height of the rocket will be h(t)=−12gt2+v0tsinθ+h0 where g=9.8 m/s2 v0=86 m/s h0=0 m θ= angle formed with the vertical

That's a parabola. You'll solve that for h(tf)=0 to find the time of flight. The horizontal component of the rocket's velocity will be vx=v0cosθ. You know that x=vxtf=104 m where tf is the time of flight. You can use that relationship to write an expression for tf in terms of v0 and θ. Substitute that into the first equation and solve for θ. Once you've got the parabola figured out, you can easily find the maximum height by finding the vertex, and you've already found the duration of the flight.

Answer 2

Answer:

$a) \varphi = 86.04^o\\b) H = 1.8 m\\c) t = 1.21 s$

Explanation:

Given

initial velocity u = 86 m/s

Horizontal distance R = 104 m

Solution

a)

Horizontal range

$R = \frac{u^2sin2\theta}{g} \\\\104 = \frac{86^2sin2\theta}{9.8}\\\\104 = \frac{86^2sin2\theta}{9.8}\\\\2\theta = 7.92 or 172.08\\\\\theta = 3.96^o$

This angle is made with horizontal so the angle made with the verticle

$\varphi = 90 - \theta\\\\\varphi = 90 - 3.96\\\\\varphi = 86.04^o$

b)

Maximum height

$H = \frac{u^2sin^2 \theta }{2g} \\\\H = \frac{86^2 \times (sin3.96)^2}{2 \times 9.8} \\\\H = 1.8 m$

c)

Time of flight

$t = \frac{2usin\theta}{g} \\\\t = \frac{2 \times 86 \times sin3.96}{9.8} \\\\t = 1.21 s$