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anonymous
 one year ago
Need help
A ball is thrown directly up in the air from a height of 5 feet with an initial velocity of 60 ft/s. Ignoring air resistance, how long until the ball hits the ground? Use the formula h=16t^2+60t+5, where h is the height of the ball in feet and t is the time in seconds since it is thrown. Round your answer to the nearest tenth
A.
2.1 seconds
B.
3.8 seconds
C.
2.4 seconds
D.
1.8 seconds
anonymous
 one year ago
Need help A ball is thrown directly up in the air from a height of 5 feet with an initial velocity of 60 ft/s. Ignoring air resistance, how long until the ball hits the ground? Use the formula h=16t^2+60t+5, where h is the height of the ball in feet and t is the time in seconds since it is thrown. Round your answer to the nearest tenth A. 2.1 seconds B. 3.8 seconds C. 2.4 seconds D. 1.8 seconds

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I say it's B not sure tho

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0'm gonna give you a short way to do this Let us name our vectors! Anything going up is positive and anything coming down / acting down is negative. u = +60 ft/s g = 9.8 m/s^2 (since gravity acts downward) In this case though, it is m/s^2 and not ft/s^2....Change metres to feet and we will get 32.1522 ft/s^2 s = 5 ft (s is displacement which is vector and that is the direction the ball will be falling when coming down so we have taken into consideration direction). t = what we are looking for We can therefore use the equation: s = ut + 0.5at^2 We can modify ti to get: s = ut + 0.5gt^2 Plug in values: 5 = (60)t + 0.5(32.1522)t^2 5 = 60t  16.0761t^2 FORM INTO A QUADRATIC EXPRESSION: 16.0761t^2 + 60t + 5 = 0 Solve for t by using the quadratic equation: You get t = 0.08 s (1 d.p) and t = 3.8 s TIME CAN NEVER BE NEGATIVE, so we will use the POSITIVE value for the time Hence, the time for trajectory (movement of the ball) is 3.8 seconds
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