Question

A fighter plane moving with a speed of \[50\sqrt 2 \,m/s\] upward at an angle of \[45^\circ \] with the vertical releases a bomb when it was at a height 1000 m from the ground.
Find
a. the time of flight
b. the maximum height of the bomb above ground
A) 20s,1125m
B) 22s,1155m
C) 30s,1135m
D) 40s,1145m

Answer 1

Hint: In this solution, we will use the second and the third equation of kinematics to determine the variables asked to us. The bomb when released will act like a projectile motion whose starting point is above the ground.
Formula used: In this solution, we will use the following formula:
-Second equation of kinematics: $d = ut + \dfrac{1}{2}a{t^2}$ where $d$ is the distance covered by an object travelling with an initial velocity \[u\] , constant acceleration $a$ in time $t$
-Third equation of kinematics: ${v^2} = {u^2} + 2ad$ where $v$ is the final velocity of the object, $d$ is the distance the object travels

Complete step by step answer:
We’ve been given that the fighter plane moving at an angle of \[45^\circ \] with the vertical releases a bomb when it was at a height 1000 m from the ground.
The bomb when released will be travelling in the same direction and velocity as that of the plane. Hence it will also have a velocity of \[50\sqrt 2 \,m/s\] upward at an angle of \[45^\circ \] with the vertical.
Then to determine the time of flight we need to consider the amount of time where the bomb reaches the ground. The initial vertical velocity of the bomb in the vertical direction will be $50\sqrt 2 \times \sin 45^\circ $. So, we can write
$u = 50\sqrt 2 \times \dfrac{1}{{\sqrt 2 }}$
will be
$ \Rightarrow u = 50\,m/s$
And the acceleration experienced by the bomb will be now only due to gravity. So, using the second equation of kinematics, we can write
$ - 1000 = 50t - \dfrac{1}{2} \times 10 \times {t^2}$
Solving for $t$, we get
${t^2} - 10t - 200 = 0$
Solving the above quadratic equation, we get
$(t - 20)(t + 10) = 0$
Hence the only possible value of $t$ will be
$t = 20\,s$
Hence the time of flight will be $t = 20\,s$.
To determine the maximum height of the ground, we first find the maximum height the bomb achieves from the 1000 m mark using the third equation of kinematics as
$ - {(50)^2} = 2( - 10)(h)$
Which gives us $h = 125m$
Hence the maximum height of the bomb above the ground will be
$H = 1000 + 125 = 1125\,m$

Hence the correct choice will be option (A).

Note: While using the equation of kinematics, we should be careful about the sign convention of different quantities. Quantities measured from below our reference level are considered to be negative while the above the reference level is considered to be positive.