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EXAMPLE 6 A projectile is fired with muzzle speed 290 m/s and an angle of elevation 45° from a position 30 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, 30) and so we need to adjust the parametric equations of the trajectory by adding 30 to the expression for y. With v0 = 290 m/s, ? = 45°, and g = 9.8 m/s², we have x = 290 cos(?/4)t = 145 ?2 x y = 30 + 290 sin(?/4)t - 1/2(9.8)t² = 145 ?2 x . Impact occurs when y = 0, that is 4.9t² - 145?2t - 30 = 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 = (145?2 + ?(42,050 + 588)) / 9.8 ? 41.99 Then x ? 145?2(41.99) ? 8610.50 (rounded to the nearest whole number), so the projectile hits the ground about 86.10 x m away. The velocity of the projectile is v(t) = r'(t) = 8199 x . So its speed at impact (rounded to the nearest whole number) is |v(41.99)| = ?((145?2)² + (145?2 - 9.8 * 41.99)²) = 840 x m/s.

          EXAMPLE 6 A projectile is fired with muzzle speed 290 m/s and an angle of elevation 45° from a position 30 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, 30) and so we need to adjust the parametric equations of the trajectory by adding 30 to the expression for y. With v0 = 290 m/s, ? = 45°, and g = 9.8 m/s², we have

x = 290 cos(?/4)t = 145 ?2 x

y = 30 + 290 sin(?/4)t - 1/2(9.8)t² = 145 ?2 x .

Impact occurs when y = 0, that is 4.9t² - 145?2t - 30 = 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 = (145?2 + ?(42,050 + 588)) / 9.8 ? 41.99

Then x ? 145?2(41.99) ? 8610.50 (rounded to the nearest whole number), so the projectile hits the ground about 86.10 x m away. The velocity of the projectile is

v(t) = r'(t) = 8199 x .

So its speed at impact (rounded to the nearest whole number) is
|v(41.99)| = ?((145?2)² + (145?2 - 9.8 * 41.99)²) = 840 x m/s.

EXAMPLE 6 A projectile is fired with muzzle speed 290 m/s and an angle of elevation 45° from a position 30 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, 30) and so we need to adjust the parametric equations of the trajectory by adding 30 to the expression for y. With v0 = 290 m/s, ? = 45°, and g = 9.8 m/s², we have

x = 290 cos(?/4)t = 145 ?2 x

y = 30 + 290 sin(?/4)t - 1/2(9.8)t² = 145 ?2 x .

Impact occurs when y = 0, that is 4.9t² - 145?2t - 30 = 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 = (145?2 + ?(42,050 + 588)) / 9.8 ? 41.99

Then x ? 145?2(41.99) ? 8610.50 (rounded to the nearest whole number), so the projectile hits the ground about 86.10 x m away. The velocity of the projectile is

v(t) = r'(t) = 8199 x .

So its speed at impact (rounded to the nearest whole number) is
|v(41.99)| = ?((145?2)² + (145?2 - 9.8 * 41.99)²) = 840 x m/s.

Added by Diana C.

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