Physics based Calculus

Bob has just finished climbing a sheer cliff above a beach, and wants to figure out how high he climbed. All he has to use, however, is a baseball, a stopwatch, and a friend on the ground below with a long measuring tape. Bob is a pitcher, and knows that the fastest he can throw the ball is 86.0 mph. Bob starts the stopwatch as he throws the ball (with no way to measure the ball\'s initial trajectory), and watches carefully. The ball rises then and then falls, and after 0.910 seconds the ball is once again level with Bob. Bob can\'t see well enough to time when the ball hits the ground. Bob\'s friend then measures that the ball landed 453 ft from the base of the cliff. How high up is Bob, if the ball started from exactly 5 ft above the edge of the cliff?

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1. height of ball above cliff is

h = 5 + vy*t - 16t^2

assuming vy is y-component of maximum speed of 86 mph = 126.13 ft/s

h = 0 after .910 sec, so
vy = 9.1 ft/s

now, vy^2 + vx^2 = v^2
9.1^2 + vx^2 = 126.13^2
vx = 125.8 ft/s

ball fell from cliff height
453/125.8 = 3.6 sec

after falling back to cliff height, the distance fallen to ground was

9.1t + 16t^2 = 241 ft

so, the cliff is 241 ft high

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