it hits the ground when the height is zero, right?
So just solve
-4.9t^2 + 6t + 0.6 = 0
using the quadratic formula is probably easiest.
The flight of a ball hit from a tree that is 0.6m tall can be modelled by the function h(t)=-4.9t+6t+0.6, where h(t) is the height in the metres at time t seconds. How long will it take for the ball to hit the ground?
2 answers
Quadratic Formula -b±√(b²-4ac) /(2a)
a = -4.9
b= 6
c=0.6
-6±√6²-4(-4.9)(0.6) /(2)(-4.9)
= -6±√36+11.76 /-9.8
= -6±√47.76 /-9.8
= -6+ 6.9/ -9.8, -6- 6.9/ -9.8
x = -0.09, 1.31s
Therefore, it will take the ball 1.31 seconds to hit the ground.
*If the question specifies rounding to two decimal places,
x = 1.32s
a = -4.9
b= 6
c=0.6
-6±√6²-4(-4.9)(0.6) /(2)(-4.9)
= -6±√36+11.76 /-9.8
= -6±√47.76 /-9.8
= -6+ 6.9/ -9.8, -6- 6.9/ -9.8
x = -0.09, 1.31s
Therefore, it will take the ball 1.31 seconds to hit the ground.
*If the question specifies rounding to two decimal places,
x = 1.32s