Answer:
a. 4.5secs
Explanation:
From the question, the equation describing the height is given by
[tex]h(t)=320-16t^{2}\\[/tex]
at the point when the rock hit the ground, the height,h will be zero.
Hence we can have
[tex]h(t)=320-16t^{2}\\\\at h(t)=0\\0=320-16t^{2}\\hence \\320=16t^{2}\\t^{2}=\frac{320}{16}\\ t^{2}=20\\t=\sqrt{20}\\ t=4.47secs\\t=4.5sec\\[/tex]
hence the rock will hit the ground in 4.5secs
