Question

Frim the castle wall 20 m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above horizontal. Find the range and hI e final velocity

Answer 1

Range of arrow = 225.09 meter

Final horizontal velocity = 34.47 m/s

Step-by-step explanation:

We have equation of motion
`s=ut+(1)/(2) at^2`, where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.

Considering the vertical motion of arrow ( up direction as positive)

We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity =
`-9.8m/s^2`.

`-20=28.93*t-(1)/(2) *9.8*t^2\\ \\ 4.9t^2-28.93t-20=0`

t = 6.53 seconds or t = -0.63 seconds

So time = 6.53 seconds.

Considering the horizontal motion of arrow

u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a =
`0m/s^2`

`s=34.47*6.53+(1)/(2) *0*6.53^2\\ \\ s=225.09m`

So range of arrow = 225.09 meter

Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.