Question

If you decide to launch a ball vertically to a friend located 45 m above you who can catch it. What is the minimum launch speed you can use?
A) 4.5 m/s
B) 1.50 m/s
C) 45 m/s
D) 29.7 m/s
E) 90 m/s

Answer 1

Hint
Here, as the ball is rest at finally then v = 0, and the distance is given S = 45m, we have to find the value of v, for this we will use the Newton’s equation of motion i.e. ${v^2} - {u^2} = 2gs$, g is acceleration due to gravity i.e. $9.8m{s^{ - 2}}$, on substituting the values we get the desired speed.

Complete step by step answer
Here, it is given that the ball is launched vertically to a friend and friend catch the ball therefore, it means the final velocity of is zero i.e. $v = 0m{s^{ - 2}}$
As this ball will travel the distance i.e. $S = 45m$
We have to calculate the minimum speed with which we can launch the ball.
For this, we use the Newton’s equation of motion i.e.
${v^2} - {u^2} = 2gs$ …………….. (1)
Where, v is the final speed of the ball
And u is the initial velocity of the ball
And g is acceleration due to gravity i.e. $g = 9.8m{s^{ - 2}}$
As the ball is moving vertically means against the gravity, therefore the value of gravity equal to $g = - 9.8m{s^{ - 2}}$, we have to calculate the value of u
For this, substitute the values in the equation (1), we get
$\Rightarrow - {u^2} = 2\left( { - g} \right)S$
$\Rightarrow {u^2} = 2gS$
$\Rightarrow {u^2} = 2 \times 9.8 \times 45$
$\Rightarrow {u^2} = 882m{s^{ - 1}}$
$\Rightarrow u = \sqrt {882} = 29.7m{s^{ - 1}}$
Hence, the minimum speed to launch the ball such that his friend will catch the ball is 29.7m/s
Thus, option (D) is correct.

Note
Here, for these types of question we use Newton’s three equation of motion i.e. ${v^2} - {u^2} = 2gs$, $S = ut + \dfrac{1}{2}g{t^2}$, $v = u + gt$, where symbols have same meanings as expressed in above question. Care must be taken in writing the value of the acceleration due to gravity i.e. when the object moves in the same direction of the gravity then we use $g = 9.8m{s^{ - 2}}$, but if the body is moving against the gravity as in the above case then we must use $g = - 9.8m{s^{ - 2}}$.