Question

A cannon and a target are $5.1km$ apart located on the same level. Find the time to hit the target. Speed of the cannon is $240m{s^{ - 1}}$ . Assume no air drag.
A.24.6s
B.25.9s
C.23s
D.21s

Answer 1

Hint : We have to first find the total time of the motion and then we have to find the range in the horizontal region. Using trigonometric identities we can correlate the values and form a quadratic equation. We can solve the quadratic equation by putting the values of the square part as single term and converting it into a quadratic equation.

Complete Step By Step Answer:
In order to solve this question first we have to calculate the total time of the motion that is given by;
$\tau = \dfrac{{2{v_0}\sin \alpha }}{g}$
Also,
$\sin \alpha = \dfrac{{\tau g}}{{2{v_0}}}$ That is equal to $\dfrac{{9.8\tau }}{{2 \times 240}}$ .
Now for getting the range in the horizontal region;
$R = {v_0}\cos \alpha \tau ,\cos \alpha = \dfrac{R}{{{v_0}\tau }} = \dfrac{{1500}}{{240\tau }}$
Finally we get it as $\dfrac{{85}}{{4\tau }}$ .
According to the trigonometric function, ${\sin ^2}\alpha + {\cos ^2}\alpha = 1$
So here we have to first square and then add both the equations.
We get;
$\dfrac{{{{(9.8\tau )}^2}}}{{{{(480)}^2}}} + \dfrac{{{{(85)}^2}}}{{{{(4\tau )}^2}}} = 1$
On simplifying the equation further we get;
${\tau ^4} - 2400{\tau ^2} + 1083750 = 0$
Taking the ${\tau ^2}$ as single term and then treating the whole equation as quadratic equation for ${\tau ^2}$ , we get;
${\tau ^2} = \dfrac{{2400 \pm \sqrt {1425000} }}{2} = \dfrac{{2400 \pm 1194}}{2}$
Here we get two values of $\tau$ .
Thus the value of $\tau$ we get is $24.55s$ .
Depending upon the angle alpha.
Since this is nearest to option A.
So A is the correct answer.

Note :
Here we use the projectile motion trajectories formulas to calculate the time and range of the motion. In order to find the quadratic equation we use the substitution method in which we have to put a certain variable in place of root and then convert it in quadratic equation form.