Larry kicked a ball at a 50 degrees angle to a direction north. It travels a horizontal distance of 20ft 9inchs before hitting the ground .The ball travels exactly 1min 18 sec after hitting the ground.
The ball hit a mud and cause dthe ball to subside 2inches.
Question: how many meters is the peak height of the ball from the ground after it was hit to the current postion of the ball?
horizontal travel:
x = Vxt = v cos50° = 0.6428v t
.6428v t = 20.75
v = 32.2806/t
y = Vy t - 16t^2
= 0.7660v t - 16t^2
= 0.7660(32.2806/t)t - 16t^2
= 24.7270 - 16t^2
y=0 when t = 1.243
Vy (1.243) - 16(1.243^2) = 0
Vy = 19.89
SO,
y = 19.89t - 16t^2
Now just find the vertex of the parabola for max height.
No idea what the mud and subsidence have to do with anything.
To find the peak height of the ball, we need to break down the motion into vertical and horizontal components.
First, let's analyze the horizontal motion. We know that the ball travels a horizontal distance of 20 ft 9 inches, which we need to convert to meters.
1 ft = 0.3048 m (conversion factor)
1 inch = 0.0254 m (conversion factor)
20 ft = 20 * 0.3048 m = 6.096 m
9 inches = 9 * 0.0254 m = 0.2286 m
Total horizontal distance = 6.096 m + 0.2286 m = 6.3246 m
Therefore, the ball travels a horizontal distance of approximately 6.3246 meters.
Next, let's calculate the time the ball took to reach the ground. We know that it took exactly 1 minute 18 seconds, which we need to convert to seconds.
1 minute = 60 seconds
1 second (already in seconds) = 1 second
1 minute 18 seconds = 1 * 60 + 18 = 78 seconds
Therefore, the ball took approximately 78 seconds to reach the ground.
Now, we can find the vertical motion of the ball by using the formula for vertical height:
h = u*t - 0.5*g*t^2
where:
h = vertical height
u = initial vertical velocity (which is 0 since the ball was kicked horizontally)
t = time
g = acceleration due to gravity (approximately -9.8 m/s^2)
We need to find the time it took for the ball to reach the ground (t) and then backtrack to find the time it took to reach the peak height.
Since the motion is symmetrical, the time taken to reach the ground is double the time taken to reach the peak.
So, the time to reach the peak = 78 seconds / 2 = 39 seconds.
Now, we can calculate the peak height:
h = 0*t - 0.5*g*t^2
h = -0.5*(-9.8)*(39^2)
h = -0.5*(-9.8)*(1521)
h = 7419 m
Therefore, the peak height of the ball from the ground after it was hit is approximately 7419 meters.