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?
You are watching: A ball is thrown into the air
When the round hits the soil h= 0
so set your equation come = o and solve because that t
Need more help???
0 - -16t^2 +20t +6 fix for "t"
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?
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
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
See more: " Act Like A Lady Think Like A Man Quotes (Author Of Think And Grow Rich)
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)
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)