A projectile is launched with an initial velocity of 25 m/s at a launch angle of 36
degrees. What is the magnitude of the projectile’s velocity when it is 7.5 m above the ground?
v0=25 m/s
θ=36° (assumed with horizontal)
vh=v0 cos(θ)
vv=v0 sin(θ)
Let v=vertical velocity at any height S from ground, and g=9.81 m/s^2
Use v²-vv²=-2gS (for vertical direction)
v²=vv²-2gS
By kinematics:
Magnitude of Velocity,V, is the vector
sum of the vertical and horizontal velocities, sqrt(v²+vh²).
Note that vh is constant with time, and
vv²+vh²=v0 (25 m/s)
V=sqrt(v²+vh²)
=sqrt(vv²-2gS+vh²)
=sqrt(v0²-2gS)
=sqrt(25²-2*9.81*7.5)
=21.86 m/s
Note that the magnitude does not depend of the angle θ.
By energies:
Initial kinetic energy = (1/2)m(v0²)
Initial potential energy = 0 (assumed)
Final potential energy = mgH (H=7.5m)
Final Kinetic energy = (1/2)mv0²-mgH
Final magnitude of velocity
=sqrt(2(kinetic energy)/m)
=sqrt(2((1/2)25^2-9.81*7.5))
=sqrt(477.85)
=21.859
as before.
To find the magnitude of the projectile's velocity when it is 7.5 m above the ground, we can use the kinematic equations and the principles of projectile motion.
First, let's decompose the initial velocity into horizontal and vertical components. Since the launch angle is 36 degrees, the horizontal component (Vx) can be calculated as:
Vx = V * cos(θ)
= 25 * cos(36°)
= 20.3 m/s
The vertical component (Vy) can be calculated as:
Vy = V * sin(θ)
= 25 * sin(36°)
= 14.3 m/s
Using the equation for vertical motion:
y = yo + Vyo * t - (0.5 * g * t^2)
where:
y = vertical displacement (7.5 m)
yo = initial vertical position (0 m)
Vyo = initial vertical velocity (Vy)
t = time
g = acceleration due to gravity (9.8 m/s^2)
Rearranging the equation, we get:
0.5 * g * t^2 - Vyo * t + (yo - y) = 0
Substituting the values, the equation becomes:
0.5 * (9.8) * t^2 - (14.3) * t + (0 - 7.5) = 0
Simplifying the equation, we have a quadratic equation:
4.9 * t^2 - 14.3 * t - 7.5 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
where:
a = 4.9
b = -14.3
c = -7.5
Solving the quadratic equation, we find two possible values for t: t1 = 2.024 s and t2 = 0.753 s. Since the object is at the specified height only once during its trajectory, we take t2 as the correct value.
Now, using the horizontal equation of motion:
x = xo + Vx * t
where:
x = horizontal displacement (unknown)
xo = initial horizontal position (0 m)
Vx = initial horizontal velocity (Vx)
t = time (0.753 s)
Substituting the values, we get:
x = 0 + 20.3 * 0.753
= 15.29 m
The magnitude of the projectile's velocity at this point can be calculated using the Pythagorean theorem:
V = √(Vx^2 + Vy^2)
Substituting the values, we have:
V = √((20.3)^2 + (14.3)^2)
= √(412.09 + 204.49)
= √616.58
= 24.8 m/s
Therefore, the magnitude of the projectile's velocity when it is 7.5 m above the ground is approximately 24.8 m/s.