Question

In a free-fall experiment, an object is dropped from a height of h = 144 feet. A camera on the ground 500 ft from the point of impact records the fall of the object (see figure). 500 ft Not down to scale (a) Find the position function that yields the height of the object at time t assuming the object is released at time t = 0. h(t) = At what time will the object reach ground level? (Round your answer to three decimal places.) t = sec (b) Find the rates of change of the angle of elevation of the camera when t = 1 and t = 2. (Round your answers to four decimal places.) 0 (1) = rad/sec e' (2) = rad/sec

Answer 1

After 5.42 second object will be at 900 und Ang с Q2 (sina uzo) free fcall) h А sus B So height after t seconds hit - 144 - 1/2 g +² An for reaching ground level hita o Tyy O 144 288 9 (285 5.421 9 5) Now from WH= 144- 19t² dh dt 1 1/2 . 9. (2t) - gt Again famo h 500 seco, do of I, dh 500 do ch 1 at conoch at vo Now at ard ha luu- 1/9 144 - 4.9 139. ACC v (250000) + 1139-112 Z [5.8.98 Coso- 500 - 0.961 518.98 do Evo (0-90). (- 9.8) - -0.016 1sec Am

Now at & 2 144 - 29 l4U 219-8) lyu 19.6 ch at - 29 - 896 124.4 P Z AC = √ (124.4)² + (506)2 515.24 coso - 500 0.97 515.24 So do di la = 2 10.9732.(-1916) 500 0.03688326 0.03709 sec Any