Since the angle is 45˚, the vertical and horizontal components of the stone’s initial velocity are equal.
Vertical = 14 * sin 45
This is approximately 10 m/s.
During the time the stone is in the air, its vertical velocity decreases at the rate of 9.8 m/s each second. Let’s use the following equation to determine the stone’s vertical velocity as it strikes the water.
vf^2 = vi^2 + 2 * a * d
a = -9.8, d is the stone’s vertical displacement.
d = final height – initial height = 0 – 20 = -20 meters
vf^2 = (14 * sin 45)^2 + 2 * -9.8 * -20
vf^2 = 490
vf = ± √490
This is approximately 22.1 m/s. Since the stone is falling, its final vertical velocity is negative. . Let’s use the following equation to determine the total time the stone is in the air.
vf = vi – a * t
√490 = 14 * sin 45 – 9.8 * t
9.8 * t = √490 + 14 * sin 45
t = (√490 + 14 * sin 45) ÷ 9.8
The total time is approximately 3.3 seconds. During this time, the stone’s horizontal velocity is constant. Let’s use the following equation to determine the range.
d = v * cos θ * t
d = 14 * cos 45 * [(√490 + 14 * sin 45) ÷ 9.8
This is approximately 32.7meters.
b)At what velocity does the stone strike the water?
Final velocity = √(Final vertical velocity^2 + Final horizontal velocity^2)
Final velocity = √(490 + 98) = √588
This is approximately 24.2 m/s.
(14 * sin 45)^2 = 98
By using the following equation, you can determine the exact number for the stone’s final vertical velocity. Then use that number to determine the time. I think is easier than solving a quadratic equation.
vf^2 = vi^2 + 2 * a * d
I hope that this helps to understand how to solve this type of problem.
Answers & Comments
Since the angle is 45˚, the vertical and horizontal components of the stone’s initial velocity are equal.
Vertical = 14 * sin 45
This is approximately 10 m/s.
During the time the stone is in the air, its vertical velocity decreases at the rate of 9.8 m/s each second. Let’s use the following equation to determine the stone’s vertical velocity as it strikes the water.
vf^2 = vi^2 + 2 * a * d
a = -9.8, d is the stone’s vertical displacement.
d = final height – initial height = 0 – 20 = -20 meters
vf^2 = (14 * sin 45)^2 + 2 * -9.8 * -20
vf^2 = 490
vf = ± √490
This is approximately 22.1 m/s. Since the stone is falling, its final vertical velocity is negative. . Let’s use the following equation to determine the total time the stone is in the air.
vf = vi – a * t
√490 = 14 * sin 45 – 9.8 * t
9.8 * t = √490 + 14 * sin 45
t = (√490 + 14 * sin 45) ÷ 9.8
The total time is approximately 3.3 seconds. During this time, the stone’s horizontal velocity is constant. Let’s use the following equation to determine the range.
d = v * cos θ * t
d = 14 * cos 45 * [(√490 + 14 * sin 45) ÷ 9.8
This is approximately 32.7meters.
b)At what velocity does the stone strike the water?
Final velocity = √(Final vertical velocity^2 + Final horizontal velocity^2)
Final velocity = √(490 + 98) = √588
This is approximately 24.2 m/s.
(14 * sin 45)^2 = 98
By using the following equation, you can determine the exact number for the stone’s final vertical velocity. Then use that number to determine the time. I think is easier than solving a quadratic equation.
vf^2 = vi^2 + 2 * a * d
I hope that this helps to understand how to solve this type of problem.
Vx = 14*cos(45) = 9.90 m/s
Vyi = 14*sin(45) = 9.90 m/s
Equation of motion
X = x0 + v0*t + 1/2*a*t^2
Horizontally:
x = vx*t
Vertically
y = hi + vyi*t - 1/2*g*t^2
From the horizontal
t = x/vx
Vertically, when the stone hits the water y = 0. Sub in t and solve for x
0 = hi + vyi*x/vx - 1/2*g*x^2/vx^2
0 = 20 + x - 0.050*x^2
Quadratic equation
X = 32.4 m .... ans A
Get the time
t = 3.3 s
Get the final vert velocity
Vyf= vyi - g*t = -22.4 m/s
Final velocity
V = sqrt(vx^2 + vyf^2) = 24.5 m/s