When an object is propelled upward at an inclination theta to the horizontal wit

Question

When an object is propelled upward at an inclination theta to the horizontal with initial speed v0, the resulting motion is called projectile motion. See the figure below. Parametric equations that model the path of the projectile, ignoring air resistance, are given by where t is time, g is the constant accelerations due to gravity (approximately 32 ft/s^2 or 9.8 m/s^2), and h is the height from which the projectile is released. A baseball is hit with an initial speed of 115 ft/s at an angle of 40 degree to the horizontal. The ball is hit at a height of 2 ft above the ground. (a) Find parametric equations that model the position of the ball as a function of the time t in seconds. (b) What is the height of the baseball after 4 seconds? (Round your answer to two decimal places.) ft (c) What horizontal distance has the ball traveled after 4 seconds? (Round your answer to two decimal places.) ft (d) How long does it take the baseball to travel x = 350 ft? (Round your answer to two decimal places.) (e) What is the height of the baseball at the time found in (d)? (Round your answer to two decimal places.) ft (f) How long is the ball in the air before it hits the ground? (Round your answer to two decimal places.) (g) How far has the baseball traveled horizontally when it hits the ground? (Round your answer to two decimal places.) ft

Explanation / Answer

b) Ht of ball after 4 seconds = y(4) = -256+460sin40+2

= 41.68 ft.

c) Horizon. dist. traveled. = x(4) = 460cos 40 = 352.38 ft.

d) Time to travel 350ft hor. = 350/115cos40 = 3.97 seconds.

e) y(3.97) = -16(3.972) +115(3.97)sin 40+2 = 295.46-252.17 = 43.28 ft.

f) It hits the ground when y =0 or

-16t2+ 73.92t+2 =0

Solving we get t =4.647 = 4.65 seconds

g) HOrizontal distance = x(4.65) = 115(4.65)cos40 = 409.64 ft.

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