Question

You and a friend each kick a football with an initial speed of 49 feet per second. Your kick is projected at an angle of 45 degrees and your friend's kick is projected at an angle of 60 degrees. About how much farther will your football travel than your friend's football?

Answer 1

d=V^2/32 sin2 theta

pretty sure we use this

Answer 2

You should draw diagrams, well

You must find the vertical and horizontal velocities of each

for first guy:



49 ft/s / sqrt(2) = 34.648 ft/s

Since it's a 45° angle, the vertical velocity is the same too

9.8 m/s^2 = 32.1522 ft/s^2

34.648 ft/s/32.1522 ft/s^2 = 1.0776s
1.0776s * 2 = 2.1553s

Finding horizontal distance traveled:

2.1553s * 34.648 ft/s = 74.6768 ft

For second guy:

sin(60) = x/49ft/s
\[(\sqrt{3}/2)*49 ft/s = x\]
Vertical velocity = 42.435 ft/s

42.435 ft/s / 32.1522 ft/s^2 = 1.3198s
1.3198s * 2 = 2.9396s

Horizontal velocity =
.5 * 49 ft/s = 24.5 ft/s

Distance traveled
24.5 ft/s * 2.9396s = 72.0202ft

Difference:

74.6768 ft - 72.0202 ft = 2.6566 ft