Question

With what speed in miles/hour ($1\,m/s=2.23\,mi/h$) must an object be thrown to reach a height of $91.5\,m$ (equivalent to one football field)? Assume negligible air resistance.
A. 94.4 mi/h
B. 84.4 mi/h
C. 74.4 mi/h
D. 94 mi/h

Answer 1

to answer this question we will first have to apply the equations of motion in a straight line . Here we will apply the third equation of motion. We will substitute the values of final velocity, acceleration and distance travelled to find the value of initial velocity. Then at last we will convert the answer into miles/hour.

Formula used:
${v^2} = {u^2} + 2as$
Where, $v$ = final velocity of the object, $u$ = initial velocity of the object, $a$ = acceleration of the body/object and $s$=height to which the object is thrown.

Complete step by step answer:
We know that the body is moving in a straight line. Hence we can apply equations of motion to solve this problem. Here we have been given: $S=91.5\,m$, acceleration=$a=g=9.8\,m/s$ (acceleration is equal to acceleration due to gravity because here we are considering vertical motion of the body, and when the body moves vertically it experiences only acceleration due to gravity and air resistance.

But here it has been given that air resistance is 0. Hence we will consider only acceleration due to gravity) also we know that when a body is thrown, initially the body will be at rest. Hence $u=0\,m/s$. Hence we will apply the third equation of motion i.e,
${v^2} = {u^2} + 2as$
Resolving in the vertical direction
And applying the equation we get:
${v^2} = {u^2} + 2as$
The acceleration due to gravity is :
$a = g = - 9.8\,m{s^{ - 2}}$
The height is:
$\;s = h = 91.5\,m$
The final velocity is
$\;v = 0\,m{s^{ - 1}}$

Therefore, substituting the values we get:
$0 = {u^2} - 2 \times 9.8 \times 91.5$
$\Rightarrow {u^2} = 2 \times 9.8 \times 91.5$
$\Rightarrow {u^2} = 1793.4$
$\Rightarrow u = \sqrt {1793.4}$
$\Rightarrow u = 42.35m{s^{ - 1}}$
$\Rightarrow u = 42.35 \times 2.23\,miles/hour$
$\Rightarrow u = 94.44\,miles/hour$

Hence the correct answer is option A.

Note: Students should not make a mistake in the sign convention for acceleration due to gravity. The sign for acceleration due to gravity for an upward moving motion is always negative and for downward movement is always positive.