Jessica attains a height of 4.7 feet above the launch and landing ramps after 1 second. Her initial velocity is 25 feet per second. Find the angle of her launch.
a. Which equation can you use with the given information to solve for theta ? b. Substitute the known values and solve for theta.
What is Jessica’s height above the launch and landing ramps after 0.5 second?
What distance has Jessica traveled after 1 second?
I am assuming the landing ramp is at the same height as the launch ramp. So, we can work as though that's y=0
y(t) = (v sinθ)t - 16t^2
So, with y(1) = 4.7,
25sinθ - 16 = 4.7
Now you can find y for any t.
The horizontal speed is a constant 25 cosθ
so, x(t) = (25cosθ)t
Vo = 25ft./s[Ao].
Xo = 25*cosA.
Yo = 25*sinA.
Y^2 = Yo^2 + 2g*h = 0.
Yo^2 +(- 64)*4.7 = 0,
Yo = 17.3 ft/s. = vert. component of initial velocity.
a. Yo = 25*sinA = 17.3 ft/s.
b. A = 43.8o = angle of her launch.
h = Yo*t + 0.5g*t^2 = 17.3*0.5 + (-16)*0.5^2 = 4.65 ft.
d = Xo * t = 25*cos43.8 * 1 =
a. To find the angle of Jessica's launch, we can use the equation for projectile motion in two dimensions. The equation is:
H = h + (v^2 * sin^2(theta)) / (2 * g)
where H is the maximum height attained, h is the initial height above the launch ramp, v is the initial velocity, theta is the launch angle, and g is the acceleration due to gravity.
b. We can substitute the known values into the equation and solve for theta. Given that H = 4.7 feet, h = 0 feet (since she starts at the same height as the launch ramp), v = 25 ft/s, and g is approximately 32.2 ft/s^2, we have:
4.7 = 0 + (25^2 * sin^2(theta)) / (2 * 32.2)
Simplifying this equation, we have:
4.7 = (625 * sin^2(theta)) / 64.4
Multiplying both sides by 64.4, we get:
300.88 = 625 * sin^2(theta)
Dividing both sides by 625, we have:
0.481408 = sin^2(theta)
Taking the square root of both sides, we find:
sin(theta) = 0.69414
To find the angle itself, we can use the inverse sine function (arcsine) on a calculator or an online tool. In this case, arcsin(0.69414) is approximately 44.59 degrees.
Therefore, the angle of Jessica's launch is approximately 44.59 degrees.
To find Jessica's height above the launch and landing ramps after 0.5 seconds, we can use the equation for vertical displacement in projectile motion. The equation is:
h = h0 + v0 * t - (1/2) * g * t^2
where h is the final height, h0 is the initial height, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity.
Given that h0 = 0 feet, v0 = 25 ft/s, t = 0.5 seconds, and g is approximately 32.2 ft/s^2, we can substitute the known values into the equation:
h = 0 + 25 * 0.5 - (1/2) * 32.2 * (0.5)^2
Simplifying this expression, we have:
h = 12.5 - 4.025
Therefore, Jessica's height above the launch and landing ramps after 0.5 seconds is approximately 8.475 feet.
To find the distance Jessica has traveled after 1 second, we can use the equation for horizontal displacement in projectile motion. The equation is:
d = v0 * t * cos(theta)
where d is the horizontal displacement, v0 is the initial velocity, t is the time, and theta is the launch angle.
Given that v0 = 25 ft/s, t = 1 second, and the launch angle is approximately 44.59 degrees, we can substitute the known values into the equation:
d = 25 * 1 * cos(44.59)
Calculating this expression, we have:
d = 17.486 feet
Therefore, Jessica has traveled approximately 17.486 feet after 1 second.