Question

3. You shoot an arrow horizontally from 1.40 m off the ground. If it lands 35.0 m away,

a. What was it's final veritcal velocity
b.
How fast did it leave your bow horizontally?
before hitting the ground?
I

Answer 1

The final velocity is 5.24 m/s

The initial speed is 65.5 m/s

Step-by-step explanation:

Horizontal Motion

When an object is launched horizontally with an initial speed v from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

`v_x=v_o`

The vertical component of the velocity changes in time t because gravity makes the object fall at increasing speed given by:

`v_y=g.t`

The maximum horizontal distance traveled by the object can be calculated as follows:

`\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}`

Another useful formula allows us to calculate the distance traveled by the object in terms of time t:

`\displaystyle y=(g.t^2)/(2)`

a.

The time taken for the arrow to reach the ground can be calculated by solving the above equation for t:

`\displaystyle t=\sqrt{(2y)/(g)}`

Since y=1.40 m:

`\displaystyle t=√(0.2857)`

t = 0.53 s

The final vertical speed is:

`v_y=9.8.(0.53)=5.24\ m/s`

The final velocity is 5.24 m/s

b)

The initial speed can be calculated by solving the following equation for v:

`\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}`

`\displaystyle v=d\cdot\sqrt{\frac {g}{2h}}`

`\displaystyle v=35\cdot\sqrt{\frac {9.8}{2(1.40)}}`

v = 65.5 m/s

The initial speed is 65.5 m/s