Question

1) A projectile is launched from the floor with an initial velocity of 5.0 m/s at an angle of 30 degrees upward of horizontal.

a) For how long is the projectile in the air, as measured by the observer who launches the projectile?

b) How far away horizontally from the launch point does the projectile land, as measured by the observer who launched it?

c) It turns out that this projectile was launched and landed inside of a vehicle that was traveling at 90% of the speed of light. It was launched exactly when both observers were at the origin. To the observer standing on the ground next to (but not in) the vehicle, how long was the projectile in the air?

d) How far from the launch point does the projectile land, as viewed from the observer outside the vehicle?

e) As viewed from the observer outside the vehicle, how far away from them does the projectile land?

f) As viewed from the observer inside the vehicle how far away are the outside observer and the projectile as it lands?

Answer 1

Intial Velocity =u= 5 m/s ,=300 , g=acceleration due to gravity = 9.8m/s2

a) time of flight = t= (2*u*sin())/ g = 2*5*sin(30)/9.8 = 0.51 seconds

b) Horizontal Range = (u2*sin(2*))/ g = (25*sin(60))/ 9.8 = 2.2 metres

c) projectile was lauched from a moving vehicle travelling with a speed of 0.9c, horizontal component of velocity u*cos() = velocity of the car = 0.9c , u *sin()= 5*sin(30) = 2.5 m/s

as the vertical component of velocity remains unchanged , t= 2*u sin()/g = 2*5*sin(30)/9.8 = 0.51 seconds

d)Horizontal range = R= u*cos()*t = u cos() * 2*u *sin()/g

here in this case u*cos() = 0.9c m/s , u sin= 2.5 m/s g=9.8m/s2

= ((2.7*108)*2.5 )/9.8 = (6.75*108)/9.8 = 0.688*108 meters