Answer:
The ball takes [tex]$t=1 \cdot 917sec$[/tex] to hit the ground.
Explanation:
• The second equation of motion represents the total distances travelled by an object in a time interval of [tex]$\Delta t$[/tex] with an initial speed of [tex]$u$[/tex] and acceleration [tex]$a$[/tex].
• To find the time, ball takes to hit the ground, use the formula: [tex]$s = ut + \frac{1}{2}a{t^2}$[/tex]
Where, [tex]$t$[/tex] is time, [tex]$u$[/tex] is initial velocity, [tex]$a$[/tex] is acceleration and[tex]$s$[/tex] is displacement.
• In this case, [tex]$a = g = 9 \cdot 8m/se{c^2}$[/tex].
• Placing the value of the given initial velocity, [tex]$u=0cm/s$[/tex] and displacement,[tex]$s = 13m$[/tex] in the above formula.
[tex]& \therefore s = ut + \frac{1}{2}g{t^2} \\& \Rightarrow 13 = 0 \cdot t + \frac{1}{2} \times 9 \cdot 8 \times {t^2} \\[/tex]
[tex]& \Rightarrow 13 = 4 \cdot 9{t^2} \\& \Rightarrow {t^2} = \frac{{13}}{{4.9}} \\& \Rightarrow t = 1 \cdot 917sec \\\end{align}\][/tex]
• Hence, ball takes [tex]$t=1 \cdot 917sec$[/tex] to hit the ground.
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