midterm_examA_F05_solutions

midterm_examA_F05_solutions - Physics 40A - Midterm Exam...

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Physics 40A - Midterm Exam Solutions - Fall 2005 - Exam A Problem 1 [25 points] A stone is projected at a cliff of height h with an initial speed of 42 m/s at an angle of 60° above the horizontal, as shown below. The stone strikes at point A , 5.5 s after it was launched. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A , and (c) the maximum height H reached above the ground. Solution: This is a projectile motion problem so the horizontal acceleration is zero and the vertical acceleration is g . We define the origin of our coordinate system at the release point with + y vertically upward and + x horizontal (toward the cliff). The x and y components of the initial velocity are v 0 x = v 0 cos 0 = 42 m/s  cos60 ° = 21.0 m/s v 0 y = v 0 sin 0 42 m/s sin60 ° 36.37 m/s (a) To find the height h we can use y ­ y 0 = v 0 y t ­ 1 2 gt 2 with y ­ y 0 = h and t = 5.5 s. This gives h 36.37 m/s 5.5 s ­ 1 2 9.8 m/s 2 5.5 s 2 h = 51.8 m (b) The horizontal motion is motion with constant velocity, so v x = v 0 x = 21.0 m/s 2 The vertical component of the velocity at impact is given by v y = v 0 y ­ = 36.37 m/s ­ 9.8 m/s 2 5.5 s =­ 17.53 m/s Therefore, the speed at impact is s = v x 2 v y 2 =  21.0 m/s 2 ­ 17.53 m/s 2 s = 27.4 m/s (c) We can use v y 2 = v 0 y 2 ­ 2 g y ­ y 0 At the maximum height v y = 0 and we have y ­ y 0 = H . Therefore, solving for H we get H = v 0 y 2 2 g = 36.37 m/s 2 2
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This note was uploaded on 04/16/2008 for the course PHYS 40 taught by Professor Ellison during the Winter '08 term at UC Riverside.

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midterm_examA_F05_solutions - Physics 40A - Midterm Exam...

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