Question

Question #d5994

Answer 1

Time needed: 16 s.

Explanation:

FULL QUESTION

A ball is projected in an angle of 30 degrees from the horizontal with a velocity of `"320 ms"^(-1)`. Find

• The time needed for it to come to the maximum height;
• The horizontal range that takes to come to the maximum height;
• The maximum horizontal range that can be achieved with this velocity (the projectile angle is unknown);
• To achieve this horizontal range, what is the angle the ball has to be projected.

So, you know the initial velocity of the ball and the angle of launch. The ball's movement can be broken down into a vertical motion and a horizontal motion.

The same can be said for the ball's initial velocity, which will have a vertical component and a horizontal component, both of these components depending on the launch angle.

`{(v_(0x) = v_0 * costheta), (v_(0y) = v_0 * sintheta):}`

where `theta` - the angle of launch, equal to `30^@` for the first part of the problem and unknown for the second part.

Vertically, the motion of the ball will be influenced by gravity. This implies that the vertical component of its initial speed will be affected by the gravitational acceleration, `g`.

At maximum height, the vertical component of the ball's velocity will be zero.

This means that you can write

`underbrace(v_f)_(color(blue)("=0")) = v_text(oy) - g * t_"up"`, where

`t_"up"` - the time needed for the ball to reach maximum height.

Plug in your values into the above equation and solve for `t_'up"`

`t_"up" = v_(oy)/g = (320cancel("m")/cancel("s") * sin(30^@))/(10cancel("m")/"s"^(cancel(2))) = "32 s" * 1/2 = color(green)("16 s")`

Horizontally, the movement of the ball is not affected by any force, which implies that the horizontal component of its initial velocity will remain constant throughout the flight.

After 16 seconds, the horizontal distance covered by the ball will be

`d = v_(0x) * t_"up"`

`d = 320"m"/cancel("s") * cos(30^@) * 16cancel("s") = "5120 m" * sqrt(3)/2 = color(green)("4434 m")`

Now for the maximum horizontal range. I don't want the answer to become too long, so I won't show you how I got to this equation (you can read more on that here: http://socratic.org/questions/maximum-horizontal-range-of-a-projectile-on-ground-level-can-be-given-at-what-an).

The ball's maximum horizontal range can be determine from the equation

`R = (2 * v_0^2)/g * sin(theta) * cos(theta)`

In your case, the expression becomes

`R = (2 * 320^2"m"^(cancel(2))/cancel("s"^2))/(10cancel("m")/cancel("s"^2)) * sin(theta) * cos(theta)`

`R = "20,480 m" * sin(theta) * cos(theta)`

The maximum value of this expression comes about when

`sin(theta) * cos(theta) = "max"`

You can solve this equation by using this trigonometric identity

`sin(theta) * cos(theta) = sin(2theta)/2`

Since the maximum value of the sin function is 1, you get that

`max(sin(2theta)/2) = 1/2 underbrace(max(sin(2theta)))_(color(blue)("=1")) = 1/2`

The maximum horizontal range will thus be

`R = 20,480 * 1/2 = color(green)("10,240 m")`

To get the angle of launch that corresponds to this maximum horizontal range, simply use the fact that you need

`sin(2theta) = 1`

`arcsin(sin(2theta)) = arcsin(1) => 2theta = pi/2 => theta = color(green)(pi/4" rad")`