Question

A fireworks rocket is 61.3m above the ground when it explodes. Directly after the explosion, one of the pieces is moving with an initial velocity of 19.4m/s at an angle of 17.6^o above the horizontal. A second piece is moving with an initial velocity of 6.1m/s at an angle of 15.6^o below the horizonontal. (a) How far does the first piece travel in the horizonal direction before landing? (b) How much farther does the first piece travel than the second piece in the horizontal direction?

Answer 1

here,

h = 61.3 m

for particle 1 ,

initial speed , u1 = 19.4 m/s

theta1 = 17.6 degree

let the time taken to reach the ground be t1

for vertical direction

h = - u1 * sin(theta1) * t1 + 0.5 * g * t1^2

61.3 = - 19.4 * sin(17.6) * t1 + 0.5 * 9.81 * t1^2

solving for t1

t1 = 4.18 s

the horizontal distance , x1 = u1 * cos(theta1) * t1

x1 = 19.4 * cos(17.6) * 4.18 m

x1 = 77.3 m

b)

for the seccond particle

initial speed , u2 = 6.1 m/s

theta1 = 15.6 degree

let the time taken to reach the ground be t2

for vertical direction

h = u2 * sin(theta2) * t2 + 0.5 * g * t2^2

61.3 = 6.1 * sin(15.6) * t2 + 0.5 * 9.81 * t2^2

solving for t2

t2 = 3.37 s

the horizontal distance , x2 = u2 * cos(theta2) * t2

x2 = 6.1 * cos(15.6) * 3.37 m

x2 = 19.8 m

the horizontal difference , x = x1 - x2 = 57.5 m