Question

While performing the projectile experiment, a group found the launch speed to be 504 cm/s. The projectile launcher was then inclined at 25 so the launch point was 26.5 cm above the table. A hoop was placed 75.0 cm away.

a) How long does did it take the projectile to travel 75.0 cm?

b) What height above the table was the hoop placed at so that the ball went through it?

c) What was the velocity (in unit vector notation) of the ball as it went through the hoop?

Answer 1

Initial velocity, U = 504 cm/s
Initial horizontal velocity, Ux = U*cos25 = 504 * cos 25 = 456.78 cm/s
Initial vertical velocity, Uy = U*sin25 = 504 * sin 25 = 213 cm/s

a) Time to reach 75 cm
In horizontal direction, distance = 75 cm
constant speed = 456.78 cm/s
time taken = 75 / 456.78 = 0.164 s

b) In vertical direction:
Uy = 213 cm/s
acceleration, g = -981 cm/s^2
time, t = 0.164 s
Displacement, s = ut + 1/2 gt^2 = (213*0.164) - (0.5 * 981 * 0.164^2) = 21.74 cm
Thus, the hoop is placed 21.74 cm above the projectile launcher.
Distance of hoop from the table = 26.5 + 21.74 = 48.24 cm

c) Final horizontal velocity, Vx = Ux = 456.78 cm/s
   Final vertical velocity, Vy = Uy + at = 213 + (-981)*(0.164) = 52.12 cm/s
   Thus, final velocity in vector notation =( 456.78 i + 52.12 j ) cm/s