Question

Suppose the stone is thrown from the same height as in the example at an angle of 30.0° degrees below the horizontal. If it strikes the ground 57.0 m away, find (a) the time of flight, (b) the initial speed, and (c) the speed and the angle of the velocity vector with respect to the horizontal at impact.

Answer 1

To find the time of flight, initial speed, and the speed and angle of the velocity vector with respect to the horizontal at impact for a stone thrown at an angle of 30.0° below the horizontal and striking the ground 57.0 m away, you can use the equations of projectile motion.

Step-by-step explanation:

To solve this problem, we can use the equations of projectile motion. Given that the stone is thrown at an angle of 30.0° below the horizontal and strikes the ground 57.0 m away, we can find the time of flight, initial speed, and the speed and angle of the velocity vector at impact.

To find the time of flight, we can use the equation for horizontal distance: d = v*cos(theta)*t. Plugging in the values of d = 57.0 m, v = initial speed, and theta = 30.0°, we can solve for t.

The initial speed v can be found using the equation for horizontal distance again: d = v*cos(theta)*t. Plugging in the values of d = 57.0 m, t = time of flight, and theta = 30.0°, we can solve for v.

To find the speed at impact, we can use the equation for vertical velocity: vy = v*sin(theta) - g*t, where g is the acceleration due to gravity. Plugging in the values for vy = 0 m/s (since the stone is at the same height at impact), v = initial speed, theta = 30.0°, and t = time of flight, we can solve for v.

To find the angle of the velocity vector with respect to the horizontal at impact, we can use the equation for the angle: tan(theta) = vy/vx, where vy is the vertical velocity at impact and vx is the horizontal velocity at impact. Plugging in the values for vy = 0 m/s (since the stone is at the same height at impact), vx = v*cos(theta) (from previous calculations), and theta = 30.0°, we can solve for theta.

Answer 2

a)
`t=\sqrt{(2h-2x*tan \ \theta)/(g) }`

b)
`v_0=(x)/(tsin \theta)`

c)
`v_f=√(v_x^2+v_y^2)`

`tan \theta=(v_y)/(v_x)`

Step-by-step explanation:

Since you are not giving the height I will state it as h, then you can just replace and find the numerical value.

The problem can be divided in vertical and horizontal components, which are independant.

To find initial velocity in both components just find them using trigonometry:

`v_(0y) = v_0sin \ \theta\\v_(0x) = v_0cos \ \theta`

`h=v_(0y)t+(1)/(2) gt^2`

for horizontal flight use:

`v_(0x)=(x)/(t)`

where x=57m

From these two equations and using the decomposition of the initial velocity you can solve for t:

`t=\sqrt{(2h-2x*tan \ \theta)/(g) }`

once you know the time, you can find the initial velocity:

`v_(0x) = (x)/(t)=v_0sin \theta\\ v_0=(x)/(tsin \theta)`

Then, you need to calculate the final velocityin both components.

In the horizontal, the velocity does not change because there is no force in this direction, while in the vertical direction:

`2gh=v_(fy)^2-v_(0y)^2\\v_(fy)=\sqrt{2gh-v_(0y)^2}\\v_(fy)=√(2gh-(v_0sin \theta)^2)\\`

To find the speed just use this formula:

`v_f=√(v_x^2+v_y^2)`

and to find the angle use:

`tan \theta=(v_y)/(v_x)`