Question

A body is thrown horizontally from the top of the tower and it strikes the ground in 3 sec at an angle of 45 degrees with the horizontal. What is the speed of projection of the body?

Answer 1

Hint: Firstly, assume the initial velocity with which the body is projected be u. Then by using the formula for velocity we can find the speed with which the body is projected. Acceleration due to gravity g =9.8 m/s.
Formula:
$T=\sqrt{\dfrac{2H}{g}}$
${{v}_{y}}={{u}_{y}}+{{a}_{y}}T$

Complete answer:
In the question we are given that the body is thrown horizontally, which means the vertical component of initial velocity \[{{u}_{y}}=0\].
So, Let us assume that the body is thrown with a horizontal velocity ${{u}_{x}}$
The trajectory of the projectile is shown in the diagram below

 In the above diagram, H is the height of the tower from which the body is thrown.
${{v}_{x}}$and ${{v}_{y}}$are the horizontal and vertical component of velocity respectively when the body strikes the ground.
Now, we know that the acceleration due to gravity along x direction is ${{a}_{x}}=0$
So, the velocity along x direction will be${{v}_{x}}={{u}_{x}}+{{a}_{x}}T$
As ${{a}_{x}}=0$,$\dfrac{{{v}_{y}}}{{{v}_{x}}}=\tan 45=1$
$\therefore {{v}_{x}}={{u}_{x}}\text{ ---- (3)}$
Now,
$\begin{align}
  & {{v}_{y}}={{u}_{y}}+{{a}_{y}}T \\
 & \Rightarrow {{v}_{y}}={{a}_{y}}T\text{ }(\because {{u}_{y}}=0) \\
 & \Rightarrow {{v}_{y}}=9.8\times 3 \\
 & \therefore {{v}_{y}}=29.4m/s \\
\end{align}$
Also,
$\begin{align}
  & \dfrac{{{v}_{y}}}{{{v}_{x}}}=\tan 45=1 \\
 & \Rightarrow {{v}_{x}}={{v}_{y}}=29.4\text{ }m/s \\
\end{align}$
From equation (3) we can say that \[{{v}_{x}}={{u}_{x}}=29.4m/s\]
Hence, the speed of projection of the body is 29.4 m/s.

Additional information:
It can also be asked to find the height of the tower ,then one must follow the below steps.
We know that only gravitational force acts on a projectile, so the acceleration due to gravity is ${{a}_{y}}=9.8m{{s}^{-2}}$ . It is given in the question that the body strikes the ground in 3 seconds. So the time of flight $T=3\operatorname{s}$.
We know the formula of time of flight of projectile motion which is given by
$\begin{align}
  & T=\sqrt{\dfrac{2H}{g}}\text{ -----(1)} \\
 & \Rightarrow H=\dfrac{g{{T}^{2}}}{2}\text{ -----(2)} \\
 & \Rightarrow H=\dfrac{9.8\times {{3}^{2}}}{2} \\
\end{align}$
$\therefore H=44.1m$
So, the height of the tower is 44.1m

Note:
One must remember the formula for range and maximum height of the projectile when a body is projected from ground. Also, acceleration due to gravity value ,g can be approximated to $10m{{s}^{-2}}$.