Arocket of total mass 3180 kg is traveling in outer space with a velocity of 115 m/s toward the sun. it wishes to alter its course by 30.0°, and can do this by firing its rockets briefly in a direction perpendicular to its original motion. if the rocket gases are expelled at a speed of 1750 m/s, how much mass must be expelled?

Let Vx = 115m/s 

in order for the angle to be 30 degrees 

tan 30 = Vy/115

Vy = 66.40 m/s 

Using conservation of momentum:

m(1750) = (3180 - m) 66.40 

1750m = 211152 – 66.40m 

1750 m + 664.40 = 211152

1816.40 m = 211152

m = 211152 / 1816.40

m = 116.25 kg would be the answer

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft