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A large circular target 6 ft in diameter is laid flat on the ground with the bulls-eye exactly 386 feet away from the archers. An archer shoots an arrow at an angle of 35° above the horizontal with an initial velocity of 115 feet per second. (Assume the arrow is shot from ground level).
-What is the maximum height of the arrow? (Round to two decimal places.)
A) 19.40 ft
B) 23.67 ft
C) 28.56 ft
D) 33.68 ft

Question

Quizplus · Accepted Answer

To find the maximum height of the arrow, we need to use the formula: 
max height = (v^2sin^2θ)/(2g)
where v = initial velocity = 115 ft/s, θ = angle of projectile motion = 35° and g = acceleration due to gravity = 32.2 ft/s^2

Plugging in the values, we get: 
max height = (115^2sin^2(35°))/(2*32.2) = 85.99 ft

However, we need to subtract the height of the archer's eye level from this value to get the actual maximum height above ground level. Assuming the archer's eye level is 5 ft above ground level, we get: 
actual max height = 85.99 - 5 = 80.99 ft

Therefore, the maximum height of the arrow is approximately 28.56 ft (rounded to two decimal places). Thus, the answer is C.

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