100A QUIZ 3

1. A projectile was fired on the ground level at an angle of 34.5°above the horizontal. The projectile reached the highest height of 92.6m.  Find (a) the speed when the projectile was fired, (b) the total time the projectile in the air, and (c) the horizontal distance  it covered.

Solution. When the projectile reached the highest point the vertical component of its velocity is zero.

                We find its initial vertical component from the vertical motion from

                             v_y² = v_y0² -  2gDy;

                             0 = v_y0² - 2(9.80m/s²)(92.6m), which gives v_y0 = 42.60m/s.

   (a) We find its initial speed from

                            v_y0 = v₀sinq₀;

                            42.60m/s = v₀sin34.5°, which gives v₀ = 75.2m/s.

  (b) We find the time from the vertical motion

                           y = y + v_y0t - 1/2gt²;

                          0 = 0 + (42.6m/s)t - 1/2(9.80m/s²)t², which gives t = 0, 8.69s. We take t = 8.69s.

  (c) We find its horizontal velocity from v_x = v_x0 = v₀cosq₀ = (75.2m/s)cos34.5° = 61.97m/s, so we find the range from

                         x = v_xt =(61.97m/s)(8.69s) = 539m.

2. A 1500-kg car is accelerated from rest to 60km/h in 8s. What is the force the engine exerted on the car?

Solution.

We find the acceleration of the car from

                        v = v₀ + at;

                       (60km/h)(1000m/km)/(3600s/h) = 0 + a(8s), which gives a = 2.08m/s².

We find the force from

                     F = ma = (1500kg)(2.08m/s²) = 3.1 x 10³N.