Question

A projectile is thrown upward so that its distance above the ground after t seconds is h(t) = - 16/? +3121. After how many seconds does it reach its maximum height? O A. 10 sec OB. 6 sec OC. 26 sec O D. 19.5 sec

Answer 1

Solution-

Consider the equation representing the height of projectile after t seconds .

$h(t)=-16t^2+312t$

At maximum height of the projectile , the differentiation of h(x) is 0.

So, h'(x) =0 implies

$\frac{\mathrm{d} }{\mathrm{d} t}(-16t^2+312t)=0$

$-16(2t)+312(1)=0$l

$\left ( \because \frac{\mathrm{d}at^n }{\mathrm{d} t}=a(nt^{n-1}) \right )$

Or -32t + 312 =0

Multiplying both sides by minus sign.

32t - 312 =0

Or 32t = 312

Or t = 312/32

Or t = 9.75 seconds

Hence, after t = 9.75 seconds ≈ 10 seconds, the projectile will reach the maximum height.