Question

the initial speed of an arrow is 15m/s at an elevation of 30° . what is the speed at the highest point of its trajectory​

Answer 1

Answer:

vertical component of velocity is zero at highest point

so, v= ucos30

v=15× 3½/2

v= 7.5 underoot 3

Answer 2

Given:

Initial speed of the arrow $=15m/s$

Elevation with respect to horizontal $=30^{o}$

To find:

Speed at the highest point of the trajectory.

Solution:

Step 1

For a projectile thrown with an initial velocity $u$ at an angle $\alpha$ with respect to the ground, the maximum height reached by the projectile is given by

$H_{max}=\frac{u^{2} .sin^{2}\alpha }{2g}$

The initial velocity of projection can be resolved into its rectangular components namely $u.cos\alpha$ horizontally and $u.sin\alpha$ vertically.

If we take a look at the vertical component of velocity, this component is responsible for the height achieved by the object.

Just like a ball, which when thrown up, with some initial velocity gets acted upon by the acceleration due to gravity in the downward direction and at the maximum height that is when the potential energy becomes maximum, The Final velocity of the object becomes zero $0m/s$.

Hence,

The value of vertical component of the velocity at the highest point of the projectile will be

$V_{v} =0m/s$

Step 2

Now,

If we consider the horizontal component of the velocity, $u.cos\alpha$, this component of velocity is responsible for making the object travel the maximum range which is given by

$R=\frac{u^{2}.sin2\alpha }{g}$

In the horizontal motion of the object, we know that since no acceleration acts on the body during the flight hence, velocity at the beginning of the projectile is exactly equal to the velocity of the projectile at the end of the flight.

Hence, the horizontal component of velocity of the object at the highest point of the projectile is

$V_{h}=u.cos\alpha$

$V_{h} =15.cos(30^{o} )$

$V_{h}=15(\frac{\sqrt{3} }{2} )$

$V_{h}=12.99m/s$

Step 3

Hence, the velocity of the projectile at the highest point will be

$V=\sqrt{V_{h} ^{2} +V_{v} ^{2} }$

$V=\sqrt{(12.99)^{2} +(0)^{2} }$

$V=\sqrt{(12.99)^{2} }$

$V=12.99m/s$

Final answer:

Hence, the velocity of the projectile at the highest point of the projectile is 13 m/s (approx).