contestada

One strategy in a snowball fight is to throw
a snowball at a high angle over level ground.
While your opponent is watching this first
snowball, you throw a second snowball at a
low angle and time it to arrive at the same
time as the first.
Assume both snowballs are thrown with
the same initial speed 15.1 m/s. The first
snowball is thrown at an angle of 69◦
above
the horizontal. At what angle should you
throw the second snowball to make it hit the
same point as the first? Note the starting and
ending heights are the same. The acceleration
of gravity is 9.8 m/s^2

How many seconds after the first snowball
should you throw the second so that they
arrive on target at the same time?
Answer in units of s.

Respuesta :

Answer:

Range formula:  R = v^2 sin (2 theta) / g

If theta = 69 deg and v = 15.1

R = 15.1^2 sin 138 / 9,8 = 15.6 m

sin 138 = .669 = sin 42

So a snowball thrown at 21 deg will travel

R = 15.1 * .669^2 / 9.8 = 15.6 m

The second snowball can be thrown at 21 deg to travel the same distance

Vx = V cos theta = 15.1 * cos 69 = 5.41     first snowball

t1 = 15.6 / 5.41 = 2.88 sec

Vx = V cos theta = 15.1 cos 21 = 14.1 m/s

t2 = 15.6 / 14.1 = 1.11 sec

Difference = t1 - t2 = 1.77 sec     time delay for second snowball