Question

The vertical height of the projectile at time t is given by y = 4t – t2 and the horizontal distance covered is given by x = 3t. what is the angle of projection with the horizontal?

Answer 1

Solution:

we use the following equations to determine the vertical and horizontal position of an object at a specific time t:

For vertical: h=Vyt–at²/2

h=4t−2t²

For horizontal x=Vxt

x=3t

V is the initial vertical velocity.

In terms of the initial velocity, the following equations are used to determine the initial vertical and horizontal velocities.

Initial vertical velocity: Vy= V sin Ф

Initial horizontal vlocity: Vx= V Cos Ф.

In these equations the angle is with respect to horizontal.

Since the angles that given in this problem are with respect to vertical, we need to reverse these equations.

Initial vertical velocity Vy= V cos Ф.

Initial horizontal velocity Vx= V sin Ф.

In this question, the initial vertical velocity is Vy=4m/s(coefficient near t in equation of motion along y - axis) and initial horizontal velocity is Vx =3m/s(coefficient near t in equation of motion along x-axis).

We have a system of equations:

V Cos Ф=4...... 1

V Sin Ф =3............. 2

Dividing 2 by 1 as follows we have :

(2)÷(1) = V SinФ / V Cos Ф = tan Ф =3/4

Ф = Tan⁻¹ (0.75) =37°

= 37°

Answer 2

Range is same for complementary angles as,

range= u²sin2ø/g,

and, sin2(90-ø)= sin(180-2ø)= sin2ø.

So, the angle for the second stone will be 30°.

y1= u²sin²(60°)/2g and y2= u² sin²(30°)/2g.

From both equations, y2= y1/3.

Read more on Brainly.in - https://brainly.in/question/3197130#readmore