To determine how long it takes for the stone to reach the ground, we need to break down its motion into horizontal and vertical components. Let's consider the vertical motion first.
Given: Initial speed (u) = 20.0 m/s Launch angle (θ) = 30.0° Height from which the stone is thrown (h) = 45.0 m
We can use the equations of motion to solve for the time it takes for the stone to reach the ground.
Vertical motion equation: h = u*sin(θ)*t - (1/2)gt^2
In this equation:
- h represents the vertical displacement (45.0 m).
- u represents the initial vertical velocity (u*sin(θ)).
- g represents the acceleration due to gravity (approximately 9.8 m/s^2).
- t represents the time taken for the stone to reach the ground (what we're trying to find).
Substituting the known values into the equation:
45.0 m = (20.0 m/s)*sin(30.0°)t - (1/2)(9.8 m/s^2)*t^2
Rearranging the equation:
(1/2)*(9.8 m/s^2)*t^2 - (20.0 m/s)*sin(30.0°)*t + 45.0 m = 0
Now we can solve this quadratic equation to find the time (t) it takes for the stone to reach the ground. Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac))/(2a)
Here: a = (1/2)*(9.8 m/s^2) = 4.9 m/s^2 b = -(20.0 m/s)*sin(30.0°) = -10.0 m/s c = 45.0 m
Substituting the values into the quadratic formula:
t = [ -(-10.0 m/s) ± sqrt((-10.0 m/s)^2 - 4*(4.9 m/s^2)(45.0 m)) ] / (2(4.9 m/s^2))
Calculating:
t = [ 10.0 m/s ± sqrt(100 m^2/s^2 - 4*(4.9 m/s^2)*(45.0 m)) ] / (9.8 m/s^2)
t = [ 10.0 m/s ± sqrt(100 m^2/s^2 - 4*4.9 m/s^2 * 45.0 m) ] / (9.8 m/s^2)
t = [ 10.0 m/s ± sqrt(100 m^2/s^2 - 4*4.9 m^2/s^2 * 45.0 m) ] / (9.8 m/s^2)
t ≈ [ 10.0 m/s ± sqrt(100 m^2/s^2 - 882 m^2/s^2) ] / (9.8 m/s^2)
t ≈ [ 10.0 m/s ± sqrt(-782 m^2/s^2) ] / (9.8 m/s^2)
As we can see, the value under the square root is negative, which means the quadratic equation has no real solutions. This implies that the stone will not reach the ground in this scenario.
It's important to note that the height from which the stone is thrown, 45.0 m, is greater than the maximum height the stone can reach with an initial speed of 20.0 m/s at an angle of 30.0°. Therefore, the stone will follow a parabolic trajectory but will not reach the ground.