Question

the position coordinates of a projectile thrown from ground are given by 12t-5t2 m and x=5t m initial speed of throw is ​

Answer 1

Answer:

The initial speed of throw is 13 m/s

Explanation:

Given that the position coordinates $(5t, 12t-5t^2)$

Let the initial velocity of projectile is u and the angle that the velocity vector makes with the horizontal is α

Then, at t=0,

the horizontal component of the velocity = $u\cos\alpha$

the vertical component of the velocity = $u\sin\alpha$

After time t, the horizontal displacement x = $u\cos\alpha \times t$

Thus, $u\cos\alpha \times t = 5t$

or, $u\cos\alpha = 5$

After time t, the vertical displacement

$y=u\sin\alpha \times t-\frac{1}{2} gt^2$

or, $y=u\sin\alpha \times t-5t^2$         (Taking g = 10 m/s²)

Thus, $u\sin\alpha \times t-5t^2 = 12t-5t^2$

⇒ $u\sin\alpha=12$

Now,

$u^2=u^2\cos^2\alpha+u^2sin^2\alpha$

⇒ $u^2=5^2+12^2$

⇒ $u^2=25+144$

⇒ $u^2=169$

⇒ u = 13

Therefore, the initial speed of throw is 13 m/s