To solve this problem, we can break it into horizontal and vertical components.
First, let's find the time it takes for the ball to hit the ground. We can use the vertical component of the initial velocity:
Initial vertical velocity (Vy) = 20 m/s * sin(30°)
= 10 m/s
Using the formula for vertical motion:
Vertical displacement (Sy) = Vy * t - (1/2) * g * t^2
Where g is the acceleration due to gravity (9.8 m/s^2) and Sy is the vertical displacement.
Since the ball starts 2m above the ground, Sy can be considered as -2m:
-2 = 10t - (1/2) * 9.8 * t^2
Simplifying the equation:
-2 = 10t - 4.9t^2
Rearranging the equation:
4.9t^2 - 10t - 2 = 0
Using the quadratic formula, we get:
t = (-(-10) ± √((-10)^2 - 4 * 4.9 * -2)) / (2 * 4.9)
t ≈ 1.17 seconds (ignoring negative value)
Now, let's find the horizontal distance traveled by the ball in this time:
Horizontal velocity (Vx) = 20 m/s * cos(30°)
= 20 * √3 / 2
= 10√3 m/s
Horizontal distance (Sx) = Vx * t
= 10√3 * 1.17
Finally, let's find how far behind the thrower the ball hits the ground by considering the horizontal distance from the wall:
Distance from wall = 3m + Sx
≈ 3 + 11.7√3
≈ 20.13m
Therefore, the ball hits the ground approximately 20.13m behind the thrower.
A ball is thrown towards a vertical wall from a point 2m above the ground and 3m from the wall. the initial velocity of the ball is 20m/s at an angle of 30deg above the horizontal. if the collision of the ball with the ball is perfectly elastic. how far behind the thrower does the ball hit the ground?
3 answers
3 meters toward the wall + 3 meters back to the thrower + distance behind thrower = 10√3 * 1.17
Apologies for the previous incorrect calculation. Let's reassess the situation.
The horizontal distance traveled by the ball in time 't' can be calculated using the horizontal component of the initial velocity:
Horizontal velocity (Vx) = 20 m/s * cos(30°)
= 20 * √3 / 2
= 10√3 m/s
Horizontal distance (Sx) = Vx * t
= 10√3 * 1.17
So far, we have found that the ball travels a horizontal distance of approximately 11.71 meters.
The total horizontal distance covered by the ball is the sum of the distances traveled towards the wall, then back to the thrower.
Since the ball was initially 3 meters from the wall, it travels 3 meters towards the wall.
Then, it travels back towards the thrower for an additional 3 meters.
Finally, it covers a horizontal distance of approximately 11.71 meters behind the thrower.
Adding up all these distances:
Total horizontal distance = 3 m towards the wall + 3 m back to the thrower + 11.71 m behind the thrower
= 17.71 meters
Therefore, the ball hits the ground approximately 17.71 meters behind the thrower.
The horizontal distance traveled by the ball in time 't' can be calculated using the horizontal component of the initial velocity:
Horizontal velocity (Vx) = 20 m/s * cos(30°)
= 20 * √3 / 2
= 10√3 m/s
Horizontal distance (Sx) = Vx * t
= 10√3 * 1.17
So far, we have found that the ball travels a horizontal distance of approximately 11.71 meters.
The total horizontal distance covered by the ball is the sum of the distances traveled towards the wall, then back to the thrower.
Since the ball was initially 3 meters from the wall, it travels 3 meters towards the wall.
Then, it travels back towards the thrower for an additional 3 meters.
Finally, it covers a horizontal distance of approximately 11.71 meters behind the thrower.
Adding up all these distances:
Total horizontal distance = 3 m towards the wall + 3 m back to the thrower + 11.71 m behind the thrower
= 17.71 meters
Therefore, the ball hits the ground approximately 17.71 meters behind the thrower.
7 answers