You have a machine that launches a brick at a speed Vo at an angle theta, above

Question

 You have a machine that launches a brick at a speed Vo at an angle theta, above the horizontal. You can use this device to measure the height of a building. Given, Vo, theta, and the distance d that the brick lands from the building, find an equation for the height h of the building in terms of Vo, theta, and d. You use this machine set at initial speed 10 meters per second and theta = 30 degrees. You find that the brick lands 18m away from the base of a building. How high is it?

Explanation / Answer

The horizontal distance traveled by the brick is d = vocos* t                                                                 ==> t = d / vo cos                                                                           = 18 / 10 cos30                                                                           = 2.0784 sec So that the time taken to reach the highest point is t' = t /2 = 1.039 s Therefore vertical displacement h = vosin * t' -0.5gt'2                                                   = 10 * sin30 * 1.039 - 0.5 * 9.8 * 1.039*1.039                                                   = -17.0 m

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