# An athlete throws a javlin with velocity of 20m/s at angle of 15 degree with the horizontal . How far the javlin would hit the ground from the point of projection?

## To find the horizontal distance the javelin will travel, we can use the equation for horizontal motion:

$x = v \cos(\theta) \times t$

where:
x = horizontal distance (unknown)
v = initial velocity of the javelin = 20 m/s
$$\theta$$ = angle with the horizontal = 15 degrees
t = time of flight (unknown)

First, we need to find the time of flight. We know that the vertical motion of the javelin is affected by gravity, and we can use the following equation to find the time it takes for the javelin to reach the ground:

$y = v \sin(\theta) \times t - \frac{1}{2}g \times t^2$

where:
y = vertical distance (0 as it hits the ground)
v = initial velocity of the javelin = 20 m/s
$$\theta$$ = angle with the horizontal = 15 degrees
t = time of flight (unknown)
g = acceleration due to gravity = 9.8 m/s^2

Since the javelin hits the ground, y = 0, so we have:

$0 = v \sin(\theta) \times t - \frac{1}{2}g \times t^2$

Simplifying the equation, we get:

$0.5 \times g \times t^2 = v \sin(\theta) \times t$

$0.5 \times 9.8 \times t^2 = 20 \times \sin(15) \times t$

$4.9 \times t^2 = 20 \times \sin(15) \times t$

Dividing both sides by t, and rearranging the equation, we get:

$4.9 \times t = 20 \times \sin(15)$

$t = \frac{20 \times \sin(15)}{4.9}$

Now we can substitute the value of t in the equation for horizontal distance:

$x = v \cos(\theta) \times t$

Substituting the given values, we have:

$x = 20 \times \cos(15) \times \frac{20 \times \sin(15)}{4.9}$

Evaluating this expression will give us the horizontal distance the javelin will travel.

## To determine how far the javelin would hit the ground, we need to determine the horizontal distance it would travel. We can break down the initial velocity into its horizontal and vertical components.

The horizontal component (Vx) of the initial velocity can be calculated using the formula:
Vx = V * cos(theta),
where V is the magnitude of the initial velocity (20 m/s) and theta is the angle of projection (15 degrees).

Vx = 20 * cos(15)
Vx ≈ 20 * 0.9659
Vx ≈ 19.318 m/s

Now, we can calculate the time taken (t) for the javelin to hit the ground using the vertical motion equation:
h = (V * sin(theta) * t) - (0.5 * g * t^2),
where h is the initial vertical height (assume the javelin is thrown from the ground level) and g is the acceleration due to gravity (9.8 m/s^2).

Since the javelin is thrown from the ground, the initial height (h) is 0.

0 = (20 * sin(15) * t) - (0.5 * 9.8 * t^2)
0 = 5 * t - 4.9 * t^2

Simplifying the equation, we have:
4.9 * t^2 - 5 * t = 0
t * (4.9 * t - 5) = 0

This equation gives two solutions: t = 0 or t = 5/4.9

Since time cannot be 0 in this scenario, we have t ≈ 1.0204 s.

Now, we can find the horizontal distance (d) using the formula:
d = Vx * t

d = 19.318 * 1.0204
d ≈ 19.718 m

Therefore, the javelin would hit the ground approximately 19.718 meters from the point of projection.

## the time in air is due to the initial vertical velocity, and gravity.

hf=hi+V*sin15*t-4.9t^2, find time t in air.

t=sqrt(20sin15/4.9)
now how far?
howfar=horizVelocity*timeinair
=20cos15*t
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