Question

An athlete long jump leaves the ground at an angle of $30^\circ $and travels 7.8m. What is the takeoff speed?
A. ${\text{ 2}}{\text{.4 m/s}}$
B. ${\text{5}}{\text{.6 m/s}}$
C. ${\text{9}}{\text{.4 m/s}}$
D. ${\text{6 m/s}}$

Answer 1

Hint: Here first of all we will find out the given data and frame the equation taking initial speed and the vertical speed along with the distance formula by placing the given data and simplifying for the required resultant value.

Complete step by step answer:
Let us assume that the take -off speed is $ = u$.
Time of flight be $ = t$.
Horizontal speed of the athlete can be given by ${v_x} = u\cos 30^\circ $.
Also, given that the range of the jump be $x = 7.8\,m$.
Now, using the distance, speed and time relation –
$t = \dfrac{x}{{{v_x}}}$
Place the values in the above expression –
$t = \dfrac{{7.8}}{{u\cos 30^\circ }}$
Also, the initial velocity of the athlete is $ = u\sin 30^\circ $.
Acceleration due to gravity is $a = - g = - 9.8\,m/{s^2}$
Here, vertical displacement is $s = 0\,m/{s^2}$.

Now, by using the displacement equation –
$s = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow 0 = ut + \dfrac{1}{2}a{t^2}$
Take out common multiples common from the above expression-
$0 = u + \dfrac{1}{2}at$
Place the values in the above expression –
$u\sin 30^\circ + \dfrac{1}{2}( - 9.8)\left( {\dfrac{{7.8}}{{u\cos 30^\circ }}} \right) = 0$
Remove common multiples from the numerator and the denominator. Also take LCM (least common multiple and simplify)
$\dfrac{{{u^2}\sin 30^\circ \cos 30^\circ - 4.9(7.8)}}{{\cos 30^\circ }} = 0$
Cross-multiply the expression, when zero is multiplied with any number it gives zero as the value.
${u^2}\sin 30^\circ \cos 30^\circ - 38.22 = 0$

Make the required term the subject and move other terms on the opposite side. When you move any term from one side to the other then the sign of the terms also changes.
${u^2}\sin 30^\circ \cos 30^\circ = 38.22$
Term multiplicative on one side if moved to the opposite side then it goes to the denominator.
${u^2} = \dfrac{{38.22}}{{\sin 30^\circ \cos 30^\circ }}$
Take square-root on both the sides of the equation –
$u = \sqrt {\dfrac{{38.22}}{{\sin 30^\circ \cos 30^\circ }}} $
Place value for $\sin {30^0} = \dfrac{1}{2}$and $\cos {30^0} = \dfrac{{\sqrt 3 }}{2}$
$u = \sqrt {\dfrac{{38.22}}{{\dfrac{1}{2} \times \dfrac{{\sqrt 3 }}{2}}}} $
Simplify the above expression –
$\therefore u = 9.39\,m/s$

Hence, option C is the correct answer.

Note: Here we have used negative acceleration since the athlete is jumping and moving away from the gravity. Be careful while simplifying the equation and finding the square-root. Be good in remembering the trigonometric different angles and its values.