SOLUTION: In 1978 Udo Beyer (East Germany) set a world record in the shot-put of 72 ft 8 in. If Beyer had projected the shot straight upward with a velocity of 30 ft/sec from a height of 5 f
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-> SOLUTION: In 1978 Udo Beyer (East Germany) set a world record in the shot-put of 72 ft 8 in. If Beyer had projected the shot straight upward with a velocity of 30 ft/sec from a height of 5 f
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Question 189576This question is from textbook Elementary and Intermediate Algebra
: In 1978 Udo Beyer (East Germany) set a world record in the shot-put of 72 ft 8 in. If Beyer had projected the shot straight upward with a velocity of 30 ft/sec from a height of 5 ft, then for what values of t would the shot be under 15 ft high? This question is from textbook Elementary and Intermediate Algebra
You can put this solution on YOUR website! In 1978 Udo Beyer (East Germany) set a world record in the shot-put of 72 ft 8 in. If Beyer had projected the shot straight upward with a velocity of 30 ft/sec from a height of 5 ft, then for what values of t would the shot be under 15 ft high?
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Gravity: -16t^2, upward velocity 30t, initial height: 5'; gives us an equation of:
:
h = -16t^2 + 30t + 5
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" for what values of t would the shot be under 15 ft high?"
-16t^2 + 30t + 5 < 15
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-16t^2 + 30t + 5 - 15 = 0
:
-16t^2 + 30t - 10 = 0
Solve this using the quadratic formula: a=-16; b=30; c=-10
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You should get two solutions
t ~ .43 sec
t ~ 1.44 sec
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We can say that shot-put was below 15 ft from 0 to .43 sec
and it was below 15 ft from 1.44 sec until it hit the ground
Find out when it hit the ground (h=0) using the equation
-16t^2 + 30t + 5 = 0
;
Use a graph for this, green line represents 15 ft; time (t) on the x axis:
It was below 15 ft from 1.44 sec to approx 2 sec when it hit the ground
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did this make sense to you?
