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

This question can be solved with the help of the equations of motion.

A) The Frisbee will remain in the air for "5.87 s".

B) The frisbee will go "29.4 m" down the range.

A)

To calculate the time, the frisbee will remain in the air, we will use the second equation of motion, for the vertical motion.

$h = v_it+\frac{1}{2}gt^2$

where,

h = height = 169.2 m

vi = initial velocity's vertical component = 0 m/s

g = acceleration due to gravity = 9.81 m/s²

t = time = ?

Therefore,

$169.2\ m = (0\ m/s)(t)+\frac{1}{2}(9.81\ m/s^2)t^2\\\\t = \sqrt{\frac{(169.2\ m)(2)}{9.81\ m/s^2}}$

t = 5.87 s

B)

Now, we will calculate the horizontal range by applying the equation for constant motion. Because the velocity in the horizontal direction will remain constant due to no air resistance

s = vt

where,

s = horizontal range = ?

v= initial velocity's horizontal component = 5 m/s

t = time = 5.87 s

Therefore,

s = (5 m/s)(5.87 s)

s = 29.4 m

Learn more about equations of motion here:

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