Question

An arrow is shot horizontally off of a tower that is 33m high. The arrow lands on the ground 31m from the base of the tower. How high above the ground was the arrow after 0.13s? What is the arrow's final velocity?

Answer 1

Have a look but check my maths!

Explanation:

Let us consider the diagram:

1) Vertically we have an initial velocity `v_(yi)=0` and acceleration (downwards) equal to `g=9.81m/(s^2)` and `y_i=33` while `y_f=0`. So using: `y_f-y_i=v_(yi)t+1/2at^2`
or:
`0-33=0-(1/2)9.81t^2`
`t=2.6s`
Using: `v_(yf)=v_(yi)+at` we get:
`v_(yf)=0-9.81*2.6=-25.5m/s` (minus sign because is downwards).

2) Horizontally the velocity is `v_x=const.` (no acceleration), distance`=d=31m` and time`=t=2.6s` (from the vertical evaluation bit), so:
`v_x=d/t=31/2.6~~12m/s`

Finally :
Final velocity of arrow:
`V_f=sqrt(v_x^2+v_(yf)^2)=sqrt((12)^2+(-25.5)^2)=28.2m/s`
at an angle of `theta=tan^-1(-25.5/12)~~-65°` with minus to indicate clockwise from the positive `x` axis:

When `t=0.13s` we use:
using: `y_f-y_i=v_(yi)t+1/2at^2`
`y_f-33=0-(1/2)9.81(0.13)^2`
`y_f=32.9m`