5. You are competing in the Egg Toss Contest. Your partner tosses the egg to you

Question

5. You are competing in the Egg Toss Contest. Your partner tosses the egg to you at an initial velocity of 10 feet per second. You are standing 36 feet away. At what time () does it reach you? The Physics Prof says the formula for the egg flight is D =-16t2+ vt (D distance, v initial velocity) o=-1642 +10+7-36 6. You are surveying a lot shaped like a right triangle. You measure the short leg as x and the longer leg as x+7. The hypotenuse measures 11 feet. How long is the short side? 7. You want to puldown hard wood flooring in your den. The room measures 981 square feet. The length is 6 more than the width of the room. What is the width of the room?(ure 8. A rock is tossed offthe pierlupward 8. A rock is tossed off the pier upward at an ínitial velocity of 14 ftse.if tan initial velocityofi4ft/sec.1- the pier is 18 feet above the water, how long does it take the rock to hit the water? H(t) =-16f + vot + ho (H(t) = height, vo = initial velocity, ho-initial height) .

Explanation / Answer

5. We have D = -16t2 +vt. Here v = 10 feet per second and D = 36 feet. Then 36 = -16t2 +10t or, 16t2 -10t +36 = 0 or, on dividing both the sides by 2, we have 8t2 -5t + 18 = 0. Now, on using the quadratic formula, we have t = [-(-5) ± { (-5)2 -4*8*18}]/2*8 = [5±(25- 576)]/16 = 5/16 ± i (551)/16. This means that if,an egg is tossed with a velocity of 10 feet per second, it will not reach a distance of 36 feet.

6. We know that, as per the Pythagores theorem, the square of hypotenuse is equal to the sum of the squares of the other two sides. Here, the shorter side is x, the longer side of the right angled triangle is   x +7 and the hypotenuse is 11. Then, we have (11)2 = x2 + (x +7)2 or, 121 = x2 + x2 + 14x +49 or, 2x2 +14x – 72 = 0. On dividing both the sides by 2, we have x2 +7x -36 = 0. Then, on using the quadratic formula, we have x = [-7± { (7)2 -4*1*(-36)}]/2*1 = [ -7± (49+ 144)]/2 = (-7± 193)/2 = (-7± 13.8924)/2. Since the side of a triangle cannot be negative, we have x = (-7+13.8924)/2 = 6.8924/2 = 3.4462 = 3.45 (on rounding off to 2 decimal places). Thus, the length of the shorter side of the right angled triangle is 3.45(approx.).

7. Let the width of the room be x feet. Then the length of the room is x +6 feet. Since the area of the room is 981 sq. ft. , we have x(x+6) = 981 or, x2 +6x -981 = 0.Then, on using the quadratic formula, we have x = [ -6 ± { (6)2 -4*1*(-981)}]/2*1 = [-6 ± (36+ 3924)]/2 = [-6±3960]/2 = (-6± 62.9285)/2. Now, since the width of the room cannot be negative, we have x = (-6+62.9285)/2 or, x = 56.9285/2 = 28.46 feet ( on rounding off to 2 decimal places).Thus, the width of the room is 28.46 feet(approx.).

8. We have H(t) = -16t2 +v0 t + h0. Here, h0 = 18 feet , v0 = 14 feet per second. Now, when the stone hits the water, H(t) = 0. Therefore, 0 =-16t2 +14t +18 or, 16t2 -14t – 18 = 0. On dividing both the sides by 2, we have 8t2 -7t -9 =0 On using the quadratic formula, we have t = [ -(-7)±{ (-7)2 -4*8*(-9) }]/2*8 = [7 ± ( 49+ 288)]/16 = [7± 337]/16 = (7±18.3576)/2. Since t cannot be negative, we have t = (7 + 18.3576)/16= 25.3576/16 = 1.58485 second = 1.58 second ( on rounding off to 2 decimal paces). Thus, the stone will hit the water after 1.58 seconds(approx.).

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