A ball is hit from horizontal ground with velocity (10i+24.5j) ms-1 where the unit vectors i and j are horizontal and vertically upwards respectively.
a)State two assumptions that you should make about the ball in order to make predictions about its motion.
b)the path of the ball is shown in the diagram. (diagram: the ball goes upwards in a curve and then from its highest point downwards in a curve, underneath the curve its straight with a arrow underneath that straight line saying range)
i)show that the time of flight of the ball is 5 seconds
ii)find the range of the ball
c)in fact the ball hits a vertical wall that is 20 metres from the initial position of the ball.
(another diagram in between c and d questions showing the ground and a wall with a curve going from the ground and stopping at the wall, labelled underneath the ground from the starting point of the curve the end point of the wall is an arrow that says range.)
d)Find the height of the ball when it hits the wall.
e)If a heavier ball were projected in the same way, would your answers to part (b) of this question change?Explain why.
Please help, I'm extremely stuck with this whole difficult question and need very clear detailed explanations to help me.
i) time of flight
hf=hi+vi'*t-4.9t^2 where hf=hi=0, and vi' is 24.5j
solve for time t.
ii) range= vih*t where vih=10i
d) in the range equation, subtract 20 from the range above, then solve for time t.
hf=vi'*t-4.9t^2 solve for hf
air drag negligible
only force i gravity down so horizontal velocity is constant.
Vertical problem:
Vi up = 24.5 m/s
g down = 9.81
then
h = 0 + Vi t - 4.9 t^2
h = 0 when it hits ground
so
24.5 t = 4.9 t^2
t = 0 of course or
t = 24.5/4.9 = 5 seconds in the air
(like they said )
u = constant horizontal velocity = 10 m/s
so
range = 5 seconds * 10 m/s = 50 meters
when is x = 20 meters?
x = 20 = 10 m/s * t
t = 2 seconds
so how high is it?
h = 24.5 *2 - 4.9 (4)
= 49 - 19.6 = 29.4 meters
given our assumptions weight does not matter.
I dont understand:
hf=hi+vi'*t-4.9t^2 where hf=hi=0, and vi' is 24.5j
solve for time t.
ii) range= vih*t where vih=10i
Please write in full clear sentences so i can understand the working out
a) Two assumptions that can be made about the ball to make predictions about its motion are:
1) Negligible air resistance: Assuming that there is no significant air resistance allows us to analyze the motion of the ball as a simple projectile.
2) Uniform gravitational field: Assuming that the gravitational field is constant during the motion of the ball allows us to apply basic principles of kinematics and equations of motion.
b)
i) To find the time of flight, we can analyze the vertical motion of the ball. The ball is initially projected vertically upwards, reaches its highest point, and then falls downwards. The time taken to reach the highest point and fall back down is equal to twice the time taken to reach the highest point. Therefore, the total time of flight is given by:
Time of flight = 2 × time taken to reach highest point
The time taken to reach the highest point can be found using the equation: final velocity = initial velocity + acceleration × time
Since the ball is projected vertically upwards, the final velocity at the highest point is 0 m/s. The initial velocity in the vertical direction is 24.5 m/s (because the j-component of the initial velocity vector is 24.5 m/s) and the acceleration due to gravity is -9.8 m/s^2 (since it acts in the opposite direction to the positive j-axis).
So, using the equation mentioned above:
0 = 24.5 - 9.8 × time taken to reach highest point
Simplifying and solving for time gives:
Time taken to reach highest point = 24.5/9.8 = 2.5 s
Therefore, the time of flight of the ball is:
Time of flight = 2 × 2.5 s = 5 s
ii) To find the range of the ball, we can analyze the horizontal motion of the ball. The range is the horizontal distance traveled by the ball. The horizontal component of the initial velocity vector is 10 m/s (the i-component). Assuming no horizontal acceleration, the range (R) can be found using the equation:
Range = horizontal velocity × time of flight
Since the time of flight is already calculated as 5 seconds and the horizontal velocity is 10 m/s, we can substitute these values into the equation:
Range = 10 m/s × 5 s = 50 m
Hence, the range of the ball is 50 meters.
c) In this scenario, the ball hits a vertical wall that is 20 meters from its initial position.
d) To find the height of the ball when it hits the wall, we need to determine the vertical distance it has traveled. Since the ball was initially projected upwards and then falls downwards, the height at which it hits the wall can be calculated by finding the maximum height it reaches.
The maximum height can be calculated using the equation: final velocity^2 = initial velocity^2 + 2 × acceleration × displacement
At the highest point, the final velocity of the ball is 0 m/s (since it momentarily stops before falling back down), the initial velocity in the vertical direction is 24.5 m/s, and the acceleration due to gravity is -9.8 m/s^2 (since it acts in the opposite direction to the positive j-axis).
Substituting these values into the equation:
0 = (24.5)^2 + 2 × (-9.8) × displacement
Solving for displacement gives:
Displacement = (24.5)^2 / (2 × 9.8) = 30.81 m
Therefore, the maximum height reached by the ball is 30.81 meters. Since the ball hits the wall that is 20 meters from the initial position, the height of the ball when it hits the wall would be:
Height of the ball when it hits the wall = 30.81 - 20 = 10.81 meters
e) If a heavier ball were projected in the same way, the answers to part (b) of this question would not change, assuming that the assumptions made about negligible air resistance and a uniform gravitational field still hold. The mass of an object does not affect its horizontal motion, so the range would remain the same. The vertical motion is influenced by the gravitational force, and a heavier ball would experience a greater force due to its mass, but it would also have a greater inertia, leading to the same time of flight and maximum height reached.
a) In order to make predictions about the motion of the ball, we can make the following assumptions:
1. The ball is subject only to gravity and air resistance is negligible. This assumption neglects the effects of air resistance, which can have an impact on the motion of objects in real-life scenarios. However, in this problem, we assume air resistance does not significantly affect the ball's trajectory.
2. The ground and the wall are both horizontal, providing a consistent frame of reference for the problem. In reality, the ground and the wall might have small inclinations or irregularities that could affect the motion of the ball. However, for simplicity, we assume that they are both horizontal.
b)
i) To find the time of flight of the ball, we need to determine the time it takes for the ball to reach the highest point of its trajectory and then return to the initial level.
- The vertical component of the initial velocity is 24.5 m/s upward.
- At the highest point, the vertical velocity component reduces to zero before the ball starts to descend.
- Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can find the time it takes for the vertical velocity to become zero.
In this case, the final velocity v is 0, the initial velocity u is 24.5 m/s, and the acceleration a is the acceleration due to gravity, which is approximately -9.8 m/s^2 (negative because it acts in the opposite direction of the initial velocity).
0 = 24.5 + (-9.8)t
-24.5 = -9.8t
t = -24.5 / -9.8
t = 2.5 s
The time it takes for the ball to reach the highest point is 2.5 seconds. Since the ball takes the same amount of time to come back to the initial level, the total time of flight is 2.5 + 2.5 = 5 seconds.
ii) To find the range of the ball, we need to determine the horizontal distance covered by the ball.
- The horizontal component of the initial velocity is 10 m/s.
- The time of flight is 5 seconds (as found in part i).
Using the equation d = vt, where d is the distance, v is the velocity, and t is the time, we can find the range.
d = 10 m/s * 5 s
d = 50 meters
Therefore, the range of the ball is 50 meters.
c) Given that the ball hits a vertical wall that is 20 meters from the initial position of the ball, we need to determine the height of the ball when it hits the wall.
- The horizontal distance from the initial position to the wall is 20 meters.
- We can treat this problem as a projectile motion, with the horizontal distance being the range (as found in part ii).
- The vertical height at the wall will depend on the specific trajectory of the ball.
d) Since we know the horizontal distance (range) and the initial velocity, we can determine the vertical component of the velocity at the wall.
- The horizontal component of the initial velocity is 10 m/s.
- The time of flight is 5 seconds.
Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can find the vertical component of the velocity at the wall.
- The final velocity v is unknown, the initial velocity u is 24.5 m/s upward, the acceleration a is the acceleration due to gravity (-9.8 m/s^2), and the time t is 5 seconds.
v = 24.5 - 9.8 * 5
v = 24.5 - 49
v = -24.5 m/s
Since the vertical velocity is negative, it means the ball is descending.
To find the height of the ball when it hits the wall, we can use the equation h = ut + (1/2)at^2, where h is the height, u is the initial velocity, t is the time, and a is the acceleration due to gravity.
h = 24.5 * 5 - (1/2) * 9.8 * (5^2)
h = 122.5 - (1/2) * 9.8 * 25
h = 122.5 - 122.5
h = 0 meters
Therefore, the height of the ball when it hits the wall is 0 meters, indicating it is at the same level as the ground.
e) If a heavier ball were projected in the same way, the answers to part (b) would not change. The assumptions made in part (a) still hold true regardless of the mass of the ball. The time of flight and range depend on the initial velocity and the effects of gravity, which are independent of the mass of the object. However, the height at which the ball hits the wall (part d) could potentially change if the heavier ball behaves differently due to its mass.