Answer:
a) After 4.21 seconds
b) t=1 or t=2 seconds
Step-by-step explanation:
The height of the ball at anytime t, is modeled by:
[tex]H(t)=-16t^2+v_0t+h_0[/tex] , where [tex]v_0=64[/tex] feet per second and [tex]h_0=14[/tex] feet.
We substitute the initial velocity and the initial height to obtain:
[tex]H(t)=-16t^2+64t+14[/tex]
When the ball hit the ground, then the heigth becomes zero.
We equate the function to zero to obtain:
[tex]-16t^2+64t+14=0[/tex]
We use the quadratic formula to obtain:
[tex]t=\frac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]
where a=-16, b=64 and c=14.
We substitute to obtain:
[tex]t=\frac{-64\pm\sqrt{(-64)^2-4*-16*14} }{2*-16}[/tex]
This simplifies to:
[tex]t=2\pm\frac{78}{4}[/tex]
[tex]t=4.21[/tex] or [tex]t=-0.21[/tex]
We discard the negatively, value to get t=-4.21 seconds.
The ball will hit the ground after 4.21 seconds.
To find the time that the ball have a height of 62 feet during the descent, we set H(t)=62 and solve for t.
[tex]-16t^2+64t+14=62\\-16t^2+64t+14-62=0[/tex]
[tex]-16t^2+64t-48=0[/tex]
Divide through by 16 to get:
[tex]t^2-4t+3=0[/tex]
[tex]\implies (t-1)(t-3)=0[/tex]
[tex]\implies t=1\:or\:t=3[/tex]
There the ball will first have a height of 62 at t=1 seconds and the second time is t=3 seconds
