7. A rocket is launched at an angle of 60 degrees to the horizontal. It starts f

Question

7. A rocket is launched at an angle of 60 degrees to the horizontal. It starts from rest and moves along its initial line of motion with acceleration of "2g" for"Ti” seconds. At this time its engines are shut down and it moves as a projectile. Given IT,] Determine a. The maximum altitude reached by the rocket. Ans. HT (3+3) b. The total time of fight. Ans. T-T+3+3) c. Its horizontal range. Ans. R gT33+v3) 2 8. A particle moves in a circular path of radius "R" with constant speed. The particle makes "N" revolutions per second. Given [N, R] Determine: a. The time for the particle to complete 1 revolution. b. The speed of the particle c. The acceleration of the particle. 20 9. In an action-adventure film, the hero is supposed to throw a grenade from his car, which is moving with velocity "v", to his enemy's car, which is moving at (5/4)V,. The enemy is "d" meters in front of the hero's when he lets go of the grenade. The hero throws the grenade so that its initial velocity relative to him is 45° above th The cars are both traveling in the same direction on a level road. Given [vn, d], Determine: a. The initial velocity of the grenade relative to the hero b. The magnitude and direction of the initial velocity with respect to th c. The total horizontal distance traveled by the grenade. (Hint: Fix your coordinate system on the hero's car.) e earth horizontal distance "R" frorm the cannon. Given [R, vo] Determine at what height above the cannon the net should be placed. Ans. h = R- 10. A daredevil is shot out of a cannon at 45 to the horizontal with an initial speed"v". A net is located at a gr

Explanation / Answer

7. given rocket

launched at angle phi = 60 deg

starts from rest

moves aliong initial line for t = T1 s, a = 2g

a. after T1

h = 0.5*a*T1^2*sin(theta) = gT1^2*sin(theta)

r = gT1^2*cos(theta)

v = 2g*T1

after this, when maximum height is reached

2*g*h' = v^2*sin^2(theta) = 4g^2*T1^2*sin^2(theta)

h' = 2g*T1^2*sin^2(theta)

hmax = h + h' = gT1^2(sin(60) + 2*sin^2(60)) = gT1^2(sqrt(3/2) + 3/2)

b. total time of flight = T1 + T2

-h = v*sin(theta)*T2 - gT2^2

T2^2 - 2T1T2*sin(theta) - T1^2*sin(theta) = 0

T2 = (2T1*sin(theta) + sqrt(4T1^2*sin^2(theta) + 4T1^2*sin(theta)))/2

T2 = T1*sin(60) + sqrt(T1^2*3/4 + T1^2*sqrt(3)/2)

T = T1(1 + sqrt(3) + sqrt(3 + sqrt(3)))

c. horizontal range = gT1^2*cos(theta) + 2g*T1*cos(60)*T1(sqrt(3) + sqrt(3 + sqrt(3)))

R = gT1^2[1/2 + sqrt(3) + sqrt(3 + sqrt(3)) ]

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