Answer:
The ball will reach the ground in 3.92 seconds.
Step-by-step explanation:
Given:
Height from where ball was dropped [tex](h_0) = 246\ ft[/tex]
Equation representing the height after t second.
[tex]h(t) = -16t^2+h_0[/tex]
we need to find the time (t) required by ball to reach the ground.
Solution:
To find the value of 't' when [tex]h(t) =0[/tex], presuming the ground to be at a height of zero.
Substituting the value of [tex]h(t) =0[/tex] in above equation we get;
[tex]-16t^2+246=0\\\\16t^2=246[/tex]
Dividing both side by 16 we get;
[tex]\frac{16t^2}{16}=\frac{246}{16}\\\\t^2 = 15.38[/tex]
Now taking square root on both side we get;
[tex]\sqrt{t^2} =\sqrt{15.38} \\\\t =3.921\ secs[/tex]
Rounding to 2 decimal places we get;
t = 3.92 secs
Hence The ball will reach the ground in 3.92 seconds.
It takes 3.92 seconds to reach ground level.
Given that the ball is dropped from 246 ft above the ground.
It means [tex]h_{0} = 246[/tex].
If the ball reaches the ground level then h=0.
Using these we solve the equation for t.
[tex]h=-16t^{2}+h_{0}\\0=-16t^{2}+246\\16t^2=246\\t^2=15.375\\t=\sqrt{15.375} \\t=3.92[/tex]
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