A sphere is thrown into the air. The height in feet the the ball have the right to be modeled by the equation h = -16t 2 + 20t + 6 where t is the moment in seconds, the sphere is in the air. Once will the ball hit the ground? how high will the round go?

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When the round hits the soil h= 0

so set your equation come = o and solve because that t

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0 - -16t^2 +20t +6 fix for "t" +12514

A sphere is thrown into the air. The height in feet the the ball can be modeled by the equation h = -16t 2 + 20t + 6 whereby t is the moment in seconds, the sphere is in the air. When will the round hit the ground? exactly how high will the ball go?   +121062

h = -16t^ 2 + 20t + 6 ....the ball will fight the ground when h = 0

0 = -16t^2 + 20t + 6 factor

0 = ( -8t - 2 ) ( 2t - 3) collection each aspect to 0

-8t - 2 = 0 and also 2t - 3 = 0

-8t = 2 2t = 3

t =-2/8 =- 1/4 sec t = 3/2 sec = 1.5 sec

Max ht will certainly be reached at < -b / 2a > sec = -20 / < 2(-16)> = 20/32 = 5/8 sec

Sub this ago ino the function

-16(5/8)^ 2 + 20(5/8) + 6 = 12.25 ft

Here"s the graph : https://www.desmos.com/calculator/1yozw3jh1c

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How high does the go? you can discover the max (or min) value of a quadratic through the formula

c-b^2/(4a)

6- (20^2)/(4(-16) = 12.25 feet max height

Time come hit ground

0 = -16t^2+20t+6 (had a typo after ~ the " 0 " in my previous answer)