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A brick is thrown upward from the top of a building at an angle of 15o to the horizontal and with an initial speed of 10 m/s. If the brick is in flight for 3.1 s, how tall is the building? 26.94 Your response differs from the correct answer by more than 10%. Double check your calculations. m 0/2 points | Previsus Answers SercP1IGE 3.P 033 My Notes Ask Your Teach A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.80 m/s. If the roof is pitched at 26.0° below the horizon and the roof edge is 3.30 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground. HINT (a) the time the baseball spends in the air (in s) 0 576 x s the horizontal distance from the roof edge to the point where the baseball lands on the ground (in m) 1 42 (b)
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Answer #1

Given data The initial velocity is vo-10 m/s The angle is θ=150 The time is t-3.1 s The vertical component of the initial velocity is V- V sin 6 10sin1:5 2.59 m/s Now we plug into the equation 2 2 --39.06 m This means the ball displaced downwards 39.06 m from its initial height in this time, therefore 39.06 m is the height of the building. The speed is v-3.8 m/s The distance is s 3.3 m The time is 2 2 3.3-3.8sin 26t(9.8)t2 2 3.3- 1.66t + 4.9t2 t 0.67 s The distance is 3.8sin 26 (0.67) 1.12 m

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