A stone is thrown from ground level at 72 m/s. Its speed when it reaches its highest point is 46 m/s.
Find the angle, above the horizontal, of the stone's initial velocity.
Answer in units of degree
Vo = 72m/s[Ao].
Xo = 46 m/s.
Xo = 72*Cos A = 46.
Cos A = 46/72 = 0.63889.
A = 50.3o
To find the angle of the stone's initial velocity, we can use the concept of projectile motion and trigonometry.
Let's define some variables:
- u: initial velocity of the stone (in m/s)
- v: final velocity of the stone (in m/s)
- θ: angle of the stone's initial velocity above the horizontal (in degrees)
We are given:
- u = 72 m/s
- v = 46 m/s
Since the projectile reaches its highest point when its vertical velocity component becomes zero, we can use the equations of motion to find the time it takes to reach the highest point:
v = u + at
0 = u + (-9.8 m/s^2)t (vertical acceleration due to gravity is -9.8 m/s^2, as it acts opposite to the upward velocity)
t = u/9.8
By substituting the given values, we can find the time it takes for the stone to reach the highest point:
t = 72/9.8 ≈ 7.35 seconds
At the highest point, the stone will have traveled half of its total time of flight. Therefore, the total time of flight is:
2t ≈ 14.7 seconds
Using this, we can find the horizontal distance traveled by the stone at the highest point:
s = ut + 0.5at^2
s = (72 m/s)(14.7 s) + 0.5(-9.8 m/s^2)(14.7 s)^2 (plug in values)
s ≈ 528.6 meters
Now, we can find the horizontal velocity component (vx) at the highest point by dividing the horizontal distance traveled by the total time of flight:
vx = s / (2t)
vx = 528.6 m / (14.7 s)
vx ≈ 35.94 m/s
Since the horizontal velocity component remains constant throughout the motion, we can use it to find the vertical velocity component (vy) at the highest point:
vy = vx * tan(θ)
Rearranging the equation, we can solve for θ:
θ = tan^(-1)(vy / vx)
Plugging in the values:
θ = tan^(-1)(46 m/s / 35.94 m/s)
Calculating this using a calculator or software, the angle θ is approximately 50.7 degrees.
Therefore, the stone's initial velocity is at an angle of approximately 50.7 degrees above the horizontal.