How long will the frisbee remain in the air? and How far down range will the frisbee go

Question

You take a frisbee to the top of the Washington Monument and send it sailing along horizontally at a speed of 5 m/s from the very top of the structure, which is 169.2 m above the ground.

A) How long will the frisbee remain in the air? (Remember--perfect physics world--no air resistance!)

 

B) How far down range will the frisbee go (down range - horizontal distance.)?

Answer 1

You need to understand 2D motion for this problem. Here are some important equations

Horizontal distance: x = V_x * t

Vertical distance: y = V_y0 * t - 0.5gt²

Horizontal velocity: (constant) V_x = V_x0

Vertical velocity: V_y = V_y0 - gt

You know the vertical distance is 169.2 m and the horizontal velocity is 5 m/s. You also know that acceleration is -9.8m/s² and the initial vertical velocity is 0 m/s

Use the vertical distance equation to find the time.

169.2 = 0t - 0.5(-9.8)(t²)

34.53 = t²

t = 5.88 seconds

To the horizontal distance, plug the time into that equation

Horizontal distance = (5)(5.88)

Horizontal distance = 29.38 m

Answer 2

Hello Erika

There are two components in this question

As the frisbee is sailing horizontally , this component of speed is not effected by gravity

so

Speed = Distance / Time

5 = Distance / Time. = X/t we have two unknown variables here

Now vertical speed of frisbee is zero and it is falling under gravity with assumed no air resistance

Using H = ut + 1/2 ( gt² ) For vertical motion

but u = 0

H= height of monument = 169.2 m= 1/2(gt² )

We can use g= 9.8 m/s²

169.2 = 4.9 t² As one equation

which gives t² = 169.2/4.9 =34.53

gives t= 5.876 sec , This is time for which it remains in air before touching the ground

Plug in this value in

5= X/t

where X= Horizontal distance covered ( in m)