In 5 second it reaches maximum height.
According to the question,
- u = 160 ft/s
- g = 32 ft/ s²
We know that,
→ [tex]s(t) = ut-(\frac{1}{2} )gt^2[/tex]
By substituting the values, we get
→ [tex]= 160t-16t^2[/tex]
Now,
→ [tex]\frac{ds}{dt} = 160-32 t[/tex]
[tex]\frac{ds}{dt} =0[/tex] at maximum height, then
→ [tex]160-32t=0[/tex]
→ [tex]32t = 160[/tex]
→ [tex]t = \frac{160}{32}[/tex]
→ [tex]= 5 \ sec[/tex]
Thus the above response is right.
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