Question

A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial speed of the projectile? (b) What is the maximum altitude? (c) What is the range? (d) Calculate the displacement from the point of launch to the position on its trajectory at 15 s.

Answer 1

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

Given:

• Time of flight = t = 20 s
• Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

Assume:

• Initial velocity of the projectile = u
• R = Range of the projectile during the time of flight
• H = maximum height of the projectile
• D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

• Initial horizontal velocity of the projectile = $u\cos \theta$
• Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed  of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

Answer 2

Projectile launched with an angle moves in projectile motion, when it is thrown in the air taking the action of gravity on it.

• (a) The initial speed of the projectile is 196 m/s.
• (b) The maximum altitude reach by the projectile is 490 meters.
• (c) The range of the projectile is 3394.82 meters.
• (d) The displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 meters.

What is projectile motion?

Projectile motion is the motion of the body, when it is thrown in the air taking the action of gravity on it.

Given information-

The initial angle of launch of a projectile is 30 degree.

The time taken by the projectile to land 20 s.

• (a) The initial speed of the projectile-

The time taken by the projectile is the time of flight for the projectile motion.

The initial speed of the projectile can be find using the distance formula of equation of motion as,

$s=u\sin(30)\times20+\dfrac{1}{2}\times9.81\times20^2$

For the same height the value of $s$ is zero. thus,
$0=u\sin(30)\times20+\dfrac{1}{2}\times9.81\times20^2\\u=196\rm m/s$

Thus the initial speed of the projectile is 196 m/s.

• (b) The maximum altitude-

As total time taken by projectile is 20 seconds. Thus to reach the maximum altitude, in the half of this time which is 10 seconds. Put this value of time in distance formula for maximum altitude as,

$h_{max}=196\sin(30)\times10+\dfrac{1}{2}\times(-9.81)\times10^2\\h_{max}=490\rm m$

The maximum altitude reach by the projectile is 490 meters.

• (c) The range-

The range along the horizontal axis can be given as,

$R=u\cos(30)t\\R=196\times\cos(30)\times20\\R=3394.82 \rm m$

Thus the range of the projectile is 3394.82 meters.

• (d) The displacement from the point of launch to the position on its trajectory at 15 s-

Height of the projectile after 15 seconds,

$h=196\sin(30)\times15+\dfrac{1}{2}\times(-9.81)\times15^2\\h=367.5\rm m$

Range of the projectile after 15 seconds,

$R=196\times \cos(30)\times15\\R=2546.11 \rm m$

Thus the displacement from the point of launch to the position on its trajectory at 15 s is,

$d=\sqrt{h^2+r^2}\\d=\sqrt{367.5^2+2546.11^2}\\d=2572.5 \rm m$

Thus, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 meters.

• (a) The initial speed of the projectile is 196 m/s.

• (b) The maximum altitude reach by the projectile is 490 meters.

• (c) The range of the projectile is 3394.82 meters.

• (d) The displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 meters.

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