a gold ball is hit with an initial velocity v0 = 53m/s at an angle of 50degrees above the horizontal.
a)how high the ball go?
b)what is the total time the ball is in the air?
c)how far will the ball travel horizontally before it hits the ground?
I will be happy to critique your thinking.
Answer for me
To answer these questions, we can break down the motion of the ball into horizontal and vertical components. Then, we can use the equations of motion to find the answers.
a) To determine how high the ball goes, we need to find the maximum height (h) reached by the ball. We can use the vertical component of the initial velocity (v0y) and the acceleration due to gravity (g = 9.8 m/s²). The equation we can use is:
v² = v₀² + 2gh
where v is the final vertical velocity at maximum height. Since the ball reaches its maximum height, the final vertical velocity (v) will be zero. Therefore, we can rearrange the equation and solve for h:
0 = v₀² + 2gh
Solving for h, we have:
h = (v₀²)/(2g)
Substituting the given values:
v₀ = 53 m/s
g = 9.8 m/s²
h = (53²)/(2 * 9.8)
h ≈ 144.85 m
Therefore, the ball goes approximately 144.85 meters high.
b) To find the total time the ball is in the air, we need to consider the vertical motion. We can use the equation:
v = v₀ - gt
where v is the final vertical velocity when the ball hits the ground. Once again, the final vertical velocity is zero. Rearranging the equation, we find:
t = v₀/g
Substituting the given values:
v₀ = 53 m/s
g = 9.8 m/s²
t = 53/9.8
t ≈ 5.41 s
Therefore, the ball is in the air for approximately 5.41 seconds.
c) To determine how far the ball will travel horizontally before hitting the ground, we can use the horizontal component of the initial velocity (v0x) and the time of flight (t) found in part b). The equation we can use is:
d = v₀x * t
where d is the horizontal distance traveled.
The horizontal component of the initial velocity (v₀x) can be found using the equation:
v₀x = v₀ * cosθ
where θ is the angle above the horizontal (50 degrees).
Substituting the given values:
v₀ = 53 m/s
θ = 50°
v₀x = 53 * cos(50°)
v₀x ≈ 34.02 m/s
Then, we can calculate the distance traveled horizontally:
d = 34.02 * 5.41
d ≈ 184.11 m
Therefore, the ball will travel approximately 184.11 meters horizontally before hitting the ground.