Question

What is the take off speed of a locust if its launch angle is 55° and its range is 0.8m?

Answer 1

Hello , you need this formula to solve. Range R = u^2 sin 2@ / g

Here R = 0.8 m; @ = 55 deg and g = 9.8 m/s^2 ; u = ?

Hence u = ./9.8*0.8/sin110 = 2.9 m/s (approx)

Answer 2

Given that,

Launch angle, $\theta=55^{\circ}$

Range of the projectile, R = 0.8 m

To find,

The take speed of the projectile.

Solution,

The range of the projectile is given by the formula as follows :

$R=\dfrac{u^2\sin2\theta}{g}$

u is take off speed

$u^2=\dfrac{Rg}{\sin2\theta}\\\\u^2=\dfrac{0.8\times 10}{\sin2(55)}\\\\u^2=8.51\\\\u=2.91\ m/s$

So, the take off speed of a locust is 2.91 m/s.

Learn more,

Projectile motion

https://brainly.in/question/16045545