A projectile from the ground has its direction of motion making an ang...
Projectile Motion
Projectile motion is a type of motion experienced by objects that are thrown, launched or dropped. The motion is two-dimensional since the object moves in both the horizontal and vertical directions. In this problem, we are given the angle of motion, the height of the object and the initial velocity. We are required to find the angle of projection.
Approach
The approach to solving this problem involves using the equations of motion for projectile motion. Since the motion is two-dimensional, we can split the motion into horizontal and vertical components. We can then use the following equations:
- Horizontal component: Vx = Vcosθ
- Vertical component: Vy = Vsinθ - gt
- Height: y = Vyt - 1/2gt^2
where V is the initial velocity, θ is the angle of projection, g is the acceleration due to gravity and t is the time.
Solution
Given:
- Initial velocity, V = 50m/s
- Height, y = 40m
- Angle of motion, θ = 45°
At the highest point of the object's motion, the vertical component of the velocity will be zero. Using the equation for the vertical component of the motion:
Vy = Vsinθ - gt
At the highest point, Vy = 0. Solving for t:
t = Vsinθ/g
Substituting the given values:
t = 50sin45°/9.8
t = 3.24s
Using the equation for the height:
y = Vyt - 1/2gt^2
Substituting the given values:
40 = (50sin45°)(3.24) - 1/2(9.8)(3.24)^2
On solving, we get:
θ = 37°
Conclusion
Therefore, the angle of projection is 37°. The projectile was launched at an angle of 37° with the horizontal and reached a height of 40m before falling back to the ground. This problem illustrates the application of the equations of motion for projectile motion and the importance of understanding the physics behind motion in order to solve problems.