Question

an arrow is shot from the top of a 475m tall building with an initial velocity of 46m/2 at an angle of 38 degrees above the horizontal. A nearby building is located 295 meters away. a) What are the horizontal and vertical components of the arrow's velocity when it reaches the nearby building? b) what are the magnitude and direction of the velocity of the arrow when it reaches the nearby building? c) what is the maximum height above the ground that the arrow reaches?

Answer 1

given height of building
h = 475 m
initial velocity v = 46 m/s
angle, theta = 38 deg above horizontal
horizontal distance from the building, d = 295 m

1. initial horizontal component of velocity = vcos(38) = 46*cos(38) = 36.248 m/s
   time taken to travel distance d = t
   vcos(38) = d/t
   t = 8.138 s

   so horizontal component of velocity after time t = 36.248 m/s
   vertical component of velocity after time t = Vy
   Vy = vsin(theta)*t - 0.5gt^2 = 46*sin(38)*8.138 - 4.9*8.138^2
   Vy = -94.0408 m/s

2. magnitude of velocity = sqroot(Vx ^2 + Vy^2) = 100.784 m/s
   direction = theta = arctan(Vy/Vx) = arctan(94.0408/36.248) = 68.92 deg below horizontal

3. maximum height abouve ground = ho
then 2*g*(ho - h) = v^2*sin^2(theta)
2*9.81*(ho - 475) = 46^2*sin^2(38)
ho = 515.87 m
so maximum height = 515.87 m