Question

Anmol fired a diwali rocket from the ground. A quadratic equation represents the height of the rocket from the ground. if the maximum height it reaches is $34$ feet after $3$ seconds it was fired, which of the following represents relation of $h$ with $t$ secs after it was fired.

• A. $h\left(t\right)=-16(t-3{)}^2+34$
• B. $h\left(t\right)=-16(t+3{)}^2+34$
• C. $h\left(t\right)=16(t-3{)}^2+34$
• D. $h\left(t\right)=16(t+3{)}^2+34$

Answer 1

Answer: (A)

$h\left(t\right)=-16(t-3{)}^2+34$

Option A: $h\left(t\right)=-16(t-3{)}^2+34$

Differentiating with respect to $t$ , $h^{\prime }\left(t\right)=-16\times 2(t-3)$

At maximum height $h^{\prime }\left(t\right)=0$. Hence $-32(t-3)=0⟹t=3$ seconds.

Also $h^{\prime \prime }\left(t\right)=-32<0$ Hence the value of $h\left(t\right)$ is maximum at $t$ $=3$.

Maximum height $=-16(3-3{)}^2+34=34$ feet.

So the conditions are satisfied and the equation is $h\left(t\right)=-16(t-3{)}^2+34$

Option B: $h\left(t\right)=-16(t+3{)}^2+34$

Differentiating with respect to $t$ , $h^{\prime }\left(t\right)=-16\times 2(t+3)$

At maximum height $h^{\prime }\left(t\right)=0$. Hence $-32(t+3)=0⟹t=-3$ seconds.

This is not possible as time is negative.

Option C: $h\left(t\right)=16(t-3{)}^2+34$

Differentiating with respect to $t$ , $h^{\prime }\left(t\right)=16\times 2(t-3)$

At maximum height $h^{\prime }\left(t\right)=0$. Hence $32(t-3)=0⟹t=3$ seconds.

$h^{\prime \prime }\left(t\right)=32>0$ Hence the value of $h\left(t\right)$ is not maximum at t $=3$.

Hence this equation doesnt satisfy the condition.

Option A: $h\left(t\right)=16(t+3{)}^2+34$

Differentiating with respect to $t$ , $h^{\prime }\left(t\right)=-16\times 2(t+3)$

At maximum height $h^{\prime }\left(t\right)=0$. Hence $-32(t+3)=0⟹t=-3$ seconds.

This is not possible as time is negative.