A projectile is fired with muzzle speed 140 m/s and an angle of elevation 45° from a position 10 m above ground level. Where does the projectile hit the ground and with what speed? SOLUTION
If we place the origin at ground level, then the initial position of the projectile is (0, 10) and so we need to adjust the parametric equations of the trajectory by adding 10 to the expression for y. With v0 = 140 m/s, α = 45°, and g = 9.8 m/s2, we have x = 140 cos(π/4)t = Correct: Your answer is correct. y = 10 + 140 sin(π/4)t − 1 2 (9.8)t2 = Correct: Your answer is correct. . Impact occurs when y = 0, that is 4.9t2 − 70 2 t − 10 = 0. Solving this quadratic equation (and using only the positive value of t), we get the following. (Round your answer to two decimal places.) t = 70 2 + 9,800 + 196 9.8 ≈ 20.30 Correct: Your answer is correct. . Then x ≈ 70 2 (20.30) ≈ 2010 Correct: Your answer is correct. (rounded to the nearest whole number), so the projectile hits the ground about 2010 Correct: Your answer is correct. m away. The velocity of the projectile is v(t) = r'(t) = . So its speed at impact (rounded to the nearest whole number) is |v(20.30)| = (70 2 )2 + (70 2 − 9.8 · 20.3