Question

Does anyone know how to do #3 and #4?

Answer 1

3. I assume that the frisby is thrown horizontally, so that its initial vertical velocity is zero. In this case, we can start by calculating the time the frisby takes to reach the bottom of the hill. The hill is h=20 m high, and the vertical motion of the frisby has a constant downward acceleration of g=-9.81 m/s^2, so the vertical position at time t is given by

`y(t)=h+(1)/(2)gt^2`

By substituting y(t)=0, we find the time t at which the frisby hits the ground:

`0=h+(1)/(2)gt^2\\t=\sqrt{(-2h)/(g)}=\sqrt{(-2(20 m))/(-9.81 m/s^2)}=2.02 s`

The horizontal motion is a uniform motion with constant speed v = 8.6 m/s, so the horizontal distance traveled by the frisby is

`d=vt=(8.6 m)(2.02 s)=17.4 m`

4. As before, we know the initial vertical position of the object (h = 122.5 m), so we can calculate the time it takes to reach the bottom of the cliff as in the previous exercise:

`0=h+(1)/(2)gt^2\\t=\sqrt{(-2h)/(g)}=\sqrt{(-2(122.5 m))/(-9.81 m/s^2)}=5.0 s`

Now we know that the object travels a distance of d=5 m in the horizontal direction, so the horizontal speed of the object is given by

`v=(d)/(t)=(5 m)/(5.0 s)=1.0 m/s`