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According to the Guinness Book of World Records, the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor-league game. The ball traveled 188m (618 ft.) before landing on the ground outside the ballpark.

Assuming that the ball's initial velocity was 41.0 degrees above the horizontal, and ignoring air resistance, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.900 m (3.00ft ) above ground level? Assume that the ground was perfectly flat.

User Akiiino
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Final answer:

To find the initial speed of the longest home run hit, we use the equations of projectile motion by breaking down the initial velocity into horizontal and vertical components and applying relevant formulas to determine the overall initial velocity required.

Step-by-step explanation:

The initial speed of a baseball required to achieve a certain projectile motion can be found using projectile motion equations. For this scenario, where a minor-league player hit a home run that traveled 188 m with an initial angle of 41 degrees, we can use the horizontal motion equation s = ut + (0.5)at^2 for distance 's' traveled, initial speed 'u', time 't', and acceleration 'a' (which is 0 in the horizontal direction if we ignore air resistance).

In the vertical direction, since the ball lands back at the same height it was hit, the vertical displacement s_y is zero. Thus, we can use 0 = (u_y)t - (0.5)g(t^2) with u_y being the initial vertical speed component, and g being the acceleration due to gravity. From these simultaneous equations, we can solve for 'u'.

Breaking down the initial speed into horizontal and vertical components, we use u_x = u cos(41) and u_y = u sin(41), where u_x must account for the 188 m distance and u_y, with gravitational acceleration factored in, results in the 0.900 m initial height being compensated over the trajectory time.

User Elazar Zadiki
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Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9

188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
User Paroxyzm
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