Question

What is a projectile? Derive the expression for the trajectory, time of flight, maximum height &

horizontal range for a projectile thrown upwards, making angle θ with the horizontal direction.
Please tell fast the answer correctly.
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Answer 1

Explanation:

explained properly by step by step..please read carefully

Answer 2

The Maximum height = H = $\frac{u^2Sin^20}{2g}$, time of flight T = 2Sinθ /g and, Horizontal Range R = Sin2θ:

Given:

A projectile is thrown upwards.

To find:

Trajectile.

Expression for projectile thrown upwards.

time of flight

maximum height and horizontal range.

Solution:

Due to an external force, the object under its force moves on its own through the air and due to gravity. This is called a projectile.

The expression first requires finding the maximum height

Initial velocity = u Sinθ

Final velocity succeeding maximum height = 0

(i) Maximum height = $v^{2} = u^2 + 2as\\0 = u^{2}Sin^2$- 2gH

H = $\frac{u^2Sin^20}{2g}$

(ii) Time of Flight = $v = u + at_1$

0 = uSinθ  - $gt_1$

$t_1 = \frac{uSin0}{g}$

t2 will be the descent:

$t_2 = t_1$ (time of descent = time of ascent)

T = t1 + t2

T = $2t_1$

Therefore time of flight T = 2Sinθ /g

(iii) Horizontal Range (R) = horizontal velocity x Time of flight

R = ucosθ T

R= ucosθ  2uSinθ /g

R = $\frac{u^2Sin0}{g}$

= 2Sinθ .Cosθ  = Sin2θ

Therefore the Maximum height = H = $\frac{u^2Sin^20}{2g}$, time of flight T = 2Sinθ /g and, Horizontal Range R = Sin2θ

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