Answer:
V ≈ 40.97 m/s
Explanation:
We can solve this problem using kinematic equations. Since the ball is thrown at an angle, we need to break down the initial velocity into its horizontal and vertical components.
The horizontal component of the velocity remains constant throughout the motion and is given by:
Vx = V₀ cosθ
where V₀ is the initial speed and θ is the angle of projection.
Substituting the given values, we have:
Vx = 28 cos(25°) ≈ 25.85 m/s
The vertical component of the velocity changes due to the acceleration of gravity. At the highest point, the vertical velocity becomes zero, and then it falls back to the ground. We can use the following kinematic equation to find the time taken for the ball to hit the ground:
Δy = V₀y t + 1/2 gt^2
where Δy is the vertical displacement, V₀y is the initial vertical velocity, g is the acceleration due to gravity (-9.8 m/s²), and t is the time taken.
Substituting the given values, we have:
-55 m = 28 sin(25°) t + 1/2 (-9.8) t^2
Simplifying, we get:
4.9 t^2 - 14 t - 55 = 0
Using the quadratic formula, we get:
t ≈ 5.95 s (rounded to two decimal places)
Therefore, the ball takes approximately 5.95 seconds to hit the ground.
To find the speed of the ball at impact, we can use the following kinematic equation:
V² = V₀² + 2gΔy
where V is the final velocity (i.e., the speed at impact).
Substituting the given values, we have:
V² = 28² - 2(-9.8)(55)
V ≈ 40.97 m/s (rounded to two decimal places)
Therefore, the speed of the ball at impact is approximately 40.97 m/s.