Question

please answare it with full details ​

Answer 1

Answer:

Given:

A projectile is thrown at 30° from 15 metre high building. Velocity of projection is 10 m/s.

To find:

Total range (distance covered) of the projectile.

Diagram:

Please refer to the attached photo to understand better.

Calculation:

We will divide the trajectory into 2 parts :

R1 and R2

R1 can be calculated as follows :

$R1 = \frac{ {u}^{2} \sin(2 \theta) }{g}$

Putting u = 10 m/s, θ = 30° , g = 10 m/s²,

$R1 = 5 \sqrt{3} \: m$

Now, R2 can be calculated as follows :

• First calculate the time taken to reach the ground.

s = ut + ½gt²

=> 15 = 5t + ½(10)t²

=> 15 = 5t + 5t²

=> t² + t - 3 = 0

Solving this Equation , we get

$t = \frac{ \sqrt{13} - 1}{2}$

$= > t = 1.3 \: seconds$

So , R2 can be calculated by Multiplying the horizontal component with time :

$R2 = 5 \sqrt{3} \times (1.3)$

$= > R2 = 11.25 \: metres$

So total distance be "d"

$d = R1 + R2$

$d = 5 \sqrt{3} + 11.25$

$= > d = 19.91 \: metres$

So final answer is 19.91 Metres.