Question

A hunter standing on a tall tower shoots an arrow into the air. The vertical component of arrows initial velocity is 10m/s and points upward. The horizontal component of arrows initial velocity is vi and points East. [If possible include a diagram, not necessary however]

a) How long does it take the arrow to reach max. height?

b) After the arrow is shot, it travels for a while and after some time the horizontal distance traveled by the arrow is 40m at which point the arrow is at the same height as the tower. What is the horizontal component of the arrows velocity (vi) right after it is shot?

c) What is the vertical component of the arrows velocity at the moment it is at the same height as the tower?

d) How high is the tower if the arrows total flight time is 4s?

Answer 1

here,

initial vertical speed , uy = 10 m/s

a)

the time taken to reach the maximum height , t = uy/g

t = 10/9.81 = 1.02 s

b)

horizontal distance , x = 40 m

the horizontal component of the arrows velocity (vi) right after it is shot , vi = x/(2*t)

vi = 38.5 m/s

c)

the final vertical velocity at the same height , vy = - uy = - 10 m/s

the final vertical velocity at the same height is 10 m/s downwards

d)

t1 = 4 s

let the height og tower be h0

for vertical direction

h0 = uy * t1 - 0.5 * g * t1^2

h0 = 10 * 4 - 0.5 * 9.81 * 4^2 m

h0 = - 38.48 m  

the height of tower is 38.48 m