+68 A ball is thrown into the air. The height in feet of the ball can be modeled by the equation h = -16t 2 + 20t + 6 where t is the time in seconds, the ball is in the air. When will the ball hit the ground? How high will the ball go?
+37146 When the ball hits the ground h= 0
so set your equation to = o and solve for t
Need more help???
0 - -16t^2 +20t +6 Solve for 't'
+12531 A ball is thrown into the air. The height in feet of the ball can be modeled by the equation h = -16t 2 + 20t + 6 where t is the time in seconds, the ball is in the air. When will the ball hit the ground? How high will the ball go?
+129845 h = -16t^ 2 + 20t + 6 ....the ball will hit the ground when h = 0
0 = -16t^2 + 20t + 6 factor
0 = ( -8t - 2 ) ( 2t - 3) set each factor to 0
-8t - 2 = 0 and 2t - 3 = 0
-8t = 2 2t = 3
t = -2/8 = - 1/4 sec [reject] t = 3/2 sec = 1.5 sec
Max ht will be reached at [ -b / 2a ] sec = -20 / [ 2(-16)] = 20/32 = 5/8 sec
Sub this back ino the function
-16(5/8)^ 2 + 20(5/8) + 6 = 12.25 ft
Here's the graph : https://www.desmos.com/calculator/1yozw3jh1c
+37146 How high does it go? You can find the max (or min) value of a quadratic by the formula
c-b^2/(4a)
6- (20^2)/(4(-16) = 12.25 feet max height
Time to hit ground
0 = -16t^2+20t+6 (had a typo after the ' 0 ' in my previous answer)
Quadratic Formula yields
t= (-20 +- sqrt(20^2 - 4(-16)(6)) / 2(-16)
= -20 +- 28 / -32 = -.25 or 1.5 sec (Throw out the neg answer....you can't have negative time)
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