Bob has just finished climbing a sheer cliff above a beach, and wants lo figure
ID: 1955317 • Letter: B
Question
Bob has just finished climbing a sheer cliff above a beach, and wants lo figure out how retch he changed 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 baa is 81. 0 mph Bob starts the stopwatch as he throws the ban (with no way to measure the ball's initial trajectory), and watches ca-equity The ball rises then and then fa-s. and after 0 910 seconds the ball is once again level with Bob. Bob cant see Ml enough lo time when the ball hits the ground Bob's friend then measures that the ball landed 431 it from the base of the diff How high up is Bob. if the ball started from easily 5 It above the edge of the cliff? Number ftExplanation / Answer
To find the initial vertical velocity of the ball, you can use: d = Vi*t + (1/2)*a*t^2 0 = Vi*(0.910) + (1/2)*(-9.8)*(0.910)^2 Vi = 4.46 m/s Using that, we can get the initial horizontal velocity of the ball: sin(angle) = Vy / V sin(angle) = 4.46 / 36.21 (equivalent to 81mph) angle = 7.08 degrees To get the horizontal velocity, you can use a similar equation: cos(angle) = Vx / V cos(7.08 degrees) = Vx / 36.21 Vx = 35.93 m/s Now, since you know the distance the ball traveled horizontally, you can find the total time the ball was in the air: D = V*t 131.37 meters (equivalent to 431 feet) = 35.93 * t t = 3.66 seconds Now, since you have the total time and the time it took to get back to the height of the cliff: d = Vi*t + (1/2)*a*t^2 d = 4.46*(3.66 - 0.910) + (1/2)*(-9.8)*(3.66 - 0.910)^2 d = -24.79 meters = -81.33 feet. The ball started 81.33 feet above the ground,so now you have to subtract 5 feet from that because the ball started 5 feet above the cliff, and you get a cliff height of 76.33 feet.