A ball is kicked from ground level so thta is intiial velocityis 10ms-1 AT AN angleof 60 degrees above the horizontal. It hits a wall which is at a distance of 8m from the initial position of the ball
a)show that the ballhits the wall 1.6s after it was kicked
b)find the heightof the ball as it hits the wall
c)find the speed of the ball as it hits the wall.
Please help with a clear explanation i have no idea
the horizontal speed is constant at 5 m/s, so it takes 8/5 = 1.6 seconds to hit the wall
the height
h(t) = 10 sin(pi/3)t - 4.9t^2
plug in t=1.6
the speed is the combination of vertical and horizontal velocities, so
vx = 10 cos(pi/3) = 5
vy = 10 sin(pi/3) - 9.8t
the speed
v^2 = (vx)^2 + (vy)^2
To solve this problem, we can break it down into different components. We'll start by finding the time it takes for the ball to hit the wall (part a), then we'll find the height of the ball (part b), and finally, we'll find the speed of the ball as it hits the wall (part c).
Let's begin with part a:
To find the time it takes for the ball to hit the wall, we need to use the horizontal motion of the ball. The initial velocity of the ball can be split into two components: the horizontal component (Vx) and the vertical component (Vy).
Vx = velocity * cos(angle)
Vy = velocity * sin(angle)
Given that the initial velocity is 10 m/s and the angle is 60 degrees, we can calculate the horizontal and vertical components:
Vx = 10 m/s * cos(60°) = 10 m/s * 0.5 = 5 m/s
Vy = 10 m/s * sin(60°) = 10 m/s * √3/2 ≈ 8.6603 m/s
Now, we can analyze the horizontal motion of the ball. The horizontal distance traveled by the ball can be calculated using the formula:
distance = velocity * time
Since the velocity is constant, we can rewrite the formula as:
distance = Vx * time
In this case, the distance is given as 8 m. Therefore, we have:
8 m = 5 m/s * time
Solving for time:
time = 8 m / 5 m/s ≈ 1.6 s
So, the ball hits the wall approximately 1.6 seconds after it was kicked.
Moving on to part b:
To find the height of the ball as it hits the wall, we can use the vertical motion of the ball. The height can be calculated using the formula:
height = Vy * time - (0.5 * acceleration * time²)
However, since we know the time it takes for the ball to reach the wall (1.6 s) and there is no vertical acceleration (assuming no air resistance), the formula simplifies to:
height = Vy * time
Plugging in the values:
height = 8.6603 m/s * 1.6 s ≈ 13.8565 m
So, the height of the ball as it hits the wall is approximately 13.8565 meters.
Finally, in part c:
To find the speed of the ball as it hits the wall, we can use the Pythagorean theorem to combine the horizontal and vertical velocities:
speed = √(Vx² + Vy²)
Plugging in the values:
speed = √(5 m/s)² + (8.6603 m/s)² ≈ √25 m²/s² + 75 m²/s² ≈ √100 m²/s² ≈ 10 m/s
Therefore, the speed of the ball as it hits the wall is approximately 10 m/s.
I hope this explanation helps you understand how to approach and solve this problem step by step. Let me know if you have any further questions!