A ball is kicked at ground level with a speed of 15.0 m/s at an angle of 35.0° to the horizontal. How much later does it hit the ground?
what is the vertical speed
15sin35.
how long is it in the air.
hf=hi+voi*t -1/2 g t^2
but hf=hi=0
heights are vertical
voi*t=1/2 g t^2
15sin35 t= 4.9t^2 solve for time t in the air.
1.1 s
To find how much later the ball hits the ground, we can use the kinematic equations of motion. Let's break down the problem using these steps:
Step 1: Resolve the initial velocity into horizontal and vertical components.
The initial speed of 15.0 m/s is at an angle of 35.0° to the horizontal. To find the horizontal and vertical components of the initial velocity, we can use trigonometry.
The horizontal component Vx = V * cos(θ)
Vx = 15.0 m/s * cos(35.0°)
Vx ≈ 15.0 m/s * 0.819 = 12.285 m/s
The vertical component Vy = V * sin(θ)
Vy = 15.0 m/s * sin(35.0°)
Vy ≈ 15.0 m/s * 0.574 = 8.61 m/s
Step 2: Find the time it takes for the ball to reach its maximum height.
When the ball reaches its maximum height, its vertical velocity becomes zero. We can use the equation:
Vy = Vy0 + gt
where Vy is the final vertical velocity, Vy0 is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time.
0 = 8.61 m/s - (9.8 m/s²)t
(9.8 m/s²)t = 8.61 m/s
t ≈ 0.878 seconds
Step 3: Find the total time the ball is in the air.
Since the ball's initial vertical velocity is positive and the final vertical velocity when it hits the ground is also positive, we can use the equation:
Vy = Vy0 + gt
0 = 8.61 m/s + (9.8 m/s²)t
(9.8 m/s²)t = -8.61 m/s
t ≈ -0.878 seconds (negative value means the ball is moving downwards)
However, time cannot be negative in this context, so we consider the positive time value.
Total time in the air = 2 * t (since the ball travels upwards and then downwards)
Total time in the air ≈ 2 * 0.878 seconds = 1.756 seconds
Hence, the ball hits the ground approximately 1.756 seconds later.