has just a sheer cliff above a beach, and wants to figure out how high he climbe

Question

has just 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 friend on the ground below with a long starts Bob and knows that the fastest he can throw the ball is 91.0 mph. Bob and watches the stopwatch as he throws the ball with Bob. Bob carefully. The ball rises and then after 0.510 seconds the is once again level the ball can't see well e to the ball hits the ground. friend then measures that ft above the landed 398 ft fromthe base of the cliff. How high up is Bob, if the ball started from exactly 5 edge of the cliff? Number 91.0 mph 0.510 s 5 ft 398 ft

Explanation / Answer

here,

initial speed of the ball , u = 91 mph

u = 40.68 m/s

let the angle be theta

t = 0.51 s

0 = u * sin(theta) - g * t/2

0 = 40.68 * sin(theta) - 9.8 * 0.51/2

theta = 3.52 degree

range , R = 398 ft

R = 119.8 m

let the time of flight be t'

u * cos(theta) * t' = 119.8

t' = 2.95 s

d = 5 ft

d = 1.5 m

and

let the height of cliff be h

- (h + d) = u * sin(theta) * t' - 0.5 * g * t'^2

- ( h + 1.5) = 40.68 *sin(3.52) 2.95 - 0.5 * 9.8 * 2.95^2

h = 33.77 m

h = 110.79 ft

the height of the cliff is 110.79 ft

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