Question

A  test  rocket  for  Cansat  is  launched  by  accelerating  it  along  a 200m inclined plane at 1.25 m/s^2 starting from rest. The incline rises  at  30  degrees  above  the  horizontal,  and  at  the  instant the rocket  leaves  it,  its  engine  turns  off  and  the  rock

et  is  now subjected  only  to  gravity  (free  fall). Along  the  ramp,  gravity  is neglected.

a)   Find  the  maximum  height  above  the  ground  that  the  rocket reaches.

b)   The  horizontal  position  of  landing  as  measured  from  the bottom of the inclined plane (initial starting point).

c)    The velocity of the rocket right before landing.

Answer 1

proff of the above questions------------------------------------------------

the net acceleration (up along) = a =1.25

v^2 (launch) = 0 + 2 a s = 2 *1.25*200 = 500

v (launch) = 22.36 m/s

height of launch = h(launch) = 200 * sin 30 = 197.60 m

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for the 2nd part:-----

angle of lanuch = 30 deg

position and velocity components after time (t)

y - h = v sin 30 * t - 0.5 gt^2 ------ (1)

x = v cos 30 * t ---- (2)

Vx = dx/dt = v cos 30 ------(3)

Vy = dy/dt = v sin 30 - gt ------ (4)

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now 3rd part:-

at max height > Vy=0 = v sin 30 - gt

t = v sin 30/g >> put in (1)

h(max) = y = h + v sin 30 *[v sin 30/g] - 0.5g[v sin 30/g]^2

h(max) = h + v^2 sin^2 (30)/2g

h(max) = 197.60 + [500*0.33/2*9.8] = 206.018 meter

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Range

for time of flight (T) >>> putting y=0 >> or rocket hits the ground

(1) >>

0 - h = v sin 30 * T - 0.5 g T^2

gT^2 - 2v sin 30 * T - 2h = 0 ------ (5)

9.8 T^2 - (30.43) T - 229.44 =0

solving >. T = 6.634 s (leaving -ve time)

R = v cos 30 * T

R = 26.53 [cos 30] *6.634

R = 144.17 meter

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Range from point A (rest) = p + R

where p = 200 cos 35 = 163.83 m

R(t) = 340 meter