A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis). The initial velocity is 60 i + 64 k, with speed measured in feet per second. The spin of the ball results in a southward acceleration of 6 ft/s2, so the acceleration vector is a = - 6 j - 32 k. Where does the ball land? (Round your answers to one decimal place.) ft from the origin at an angle of ° from the eastern direction toward the south. With what speed does the ball hit the ground? (Round your answer to one decimal place.)

The ball's position in the air at time t is given by the vector,

p (t) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ²

and its velocity is given by

v (t) = (60 i + 64 k) + (-6 j - 32 k) t

The ball is in the air for as long as it takes for the vertical (k) component of the position vector to reach 0, so we solve,

64 t - 32/2 t ² = 0 = => t = 0 OR t = 4

and so the ball is in the air for 4 s.

After this time, the ball has position vector

p (4) = (60 i + 64 k) t + 1/2 (-6 j - 32 k) t ² = 240 i - 48 j

which has magnitude

||p (4) || = √ (240² + (-48) ²) = 48 √26 ≈ 244.8 ft

in a direction θ in the x, y plane from the positive x axis such that

tanθ = - 48/240 = - 1/5 = => θ = - arctan (1/5) ≈ - 11.3º

or 11.3º South of East.

The ball hits the ground with speed

||v (4) || = ||60 i - 24 j - 64 k|| = √ (60² + (-24) ² + (-64) ²) = 4 √517 ≈ 91.0 ft/s