When I came new to As I was given these as homework after we did easy practise questions in class. I find this topic and all its worksheets easy except this one for some reason. I got every single one of these questions wrong and struggled and still do with this worksheet tough i am confident on the topic. I dont know where my working out is to right it down but I tried again in A2 and still problems. I was given a mark scheme but its just numbers like seen below and not working out or how to do this worksheet. I asked for help was emailed this sheets answers again. Please help and go through each of the questions as I have already tried them all, reviewed the mark scheme and now I need working out and equations used and why to see how this is worked out as my teacher is busy and doesnt have it. Please help.

1. A golf ball is struck at a point O on the ground and moves with an initial velocity of 20 m s–1 at an angle of 53° to the horizontal. The ball subsequently lands at a point X which is on the same horizontal level as O.

(a) Show that the time taken by the ball to reach the point X is approximately 3.26 seconds.
(b) Calculate the distance OX.
(c) State:
(i) the least speed of the ball during its flight from O to X
(ii) the direction of motion of the ball when this least speed occurs

2. A golfer hits a ball, from ground level on a horizontal surface. The initial velocity of the ball is 21 m s–1 at an angle of 60° above the horizontal. Assume that the ball is a particle and that no resistance forces act on the ball

(a) Find the maximum height of the ball
(b) Find the range of the ball
(c) Find the speed of the ball at its maximum height

3. A golf ball is hit from a position that is 4 metres higher than the horizontal area where the ball lands. The initial velocity of the ball is 30 m s–1 at an angle of 60 above the horizontal.



(a) Show that the maximum height of the ball, above the landing area, is approximately 38.4 metres.

(b) Show that the ball hits the ground approximately 5.45 seconds after it has been hit.

(c) Hence calculate the horizontal range of the ball, to the nearest metre.

4. A ball is thrown so that it initially travels at 10 m s–1 at an angle of 70° above the horizontal.

(a) A simple model assumes that the ball is thrown from ground level and lands at the same level. Find the time of flight and the range of the ball.

(b) A refined model assumes that the ball is at a height of 2 metres when it is thrown, and lands at ground level. Find the range of the ball based on this assumption.

5. A ball is kicked from ground level so that its initial velocity is 10 m s– 1, at an angle of 60° above the horizontal. It hits a wall, which is at a distance of 8 metres from the initial position of the ball.
(a) Show that the ball hits the wall 1.6 seconds after it was kicked.
(b) Find the height of the ball as it hits the wall.
(c) Find the speed of the ball as it hits the wall.

6. An arrow is fired horizontally with a speed of 30 m s–1 from a height of 2 m above ground level. Model the arrow as a particle which does not experience air resistance.
(a) Show that the time taken for the arrow to hit the ground is 0.639 seconds, correct to three significant figures.
(b) Find the horizontal distance travelled by the arrow.
(c) Describe what happens, during the flight of the arrow, to the magnitude of:
(i) the horizontal component of the velocity;
(ii) the vertical component of the velocity.

7. A ball is hit from horizontal ground with velocity (10i + 24.5j) m s–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.


(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.
(d)
(e)

Find the height of the ball when it hits the wall.

(d) If a heavier ball were projected in the same way, would your answers to part (b) of this question change? Explain why.

Solutions

1. (a) 3.26
(b) 39.2
(c) (i) 12.0 ms–1 (ii) horizontal

2. (a) 16.9m (b) 39.0m (c) 10.5 ms–1
3. (a) 38.4m
(c) 82m
4. 7.22m
5. (b) 1.31 (c) 8.62
6. (b) 19.2 m
(c) (i) Horizontal remains constant (ii) Vertical increases in magnitude
7. (a) Ball is a particle/no spin
No air resistance/Only gravity or weight
(b) (ii) 50 m (c) 29.4m
(d) No change as acceleration and initial velocity do not change with the mass

question 1. I wish I knew what your work was, as finding your erroneous thinking or math would be helpful.

<<. A golf ball is struck at a point O on the ground and moves with an initial velocity of 20 m s–1 at an angle of 53° to the horizontal. The ball subsequently lands at a point X which is on the same horizontal level as O.<<
time in air:
hf=hi+vi*sin53*t-4.9t^2
0=0+20*.798*t-4.9t^2
0=t(15.96-4.9t) or t= txt answer
horizontal distance x= vi*cos53*t= answer
So find your error. I suspect you have repeated this error in your work.

I suspect you need a tutor to sit with you as you do your thinking, and analyze your work as you do it. Can you get another student to do this and argue with you on your thinking?

To solve these physics problems, you will need to apply the principles of projectile motion and kinematics. Here's an explanation of how to solve each of the questions:

1. (a) To find the time taken by the ball to reach point X, you can use the vertical motion equation:
h = ut + (1/2)gt², where h is the vertical displacement, u is the initial vertical velocity, t is the time taken, and g is the acceleration due to gravity (-9.8 m/s²).

In this case, the initial vertical velocity u can be found using the given initial velocity and angle:
u = 20 m/s * sin(53°)

The vertical displacement h is 0 because the ball lands on the same horizontal level.
Thus, the equation becomes: 0 = (20 m/s * sin(53°))t + (1/2)(-9.8 m/s²)t².

Solve this quadratic equation for t by substituting the values and solving for t. You will find that t is approximately 3.26 seconds.

(b) To calculate the distance OX, you can use the horizontal motion equation:
s = ut + (1/2)at², where s is the horizontal displacement, u is the initial horizontal velocity, t is the time taken, and a is the acceleration (which is 0 in this case).

The initial horizontal velocity u can be found using the given initial velocity and angle:
u = 20 m/s * cos(53°)

Substitute the values into the equation and solve for s. You will find that the distance OX is approximately 39.2 meters.

(c) (i) The least speed of the ball during its flight from O to X occurs when the ball reaches the highest point of its trajectory. At this point, the vertical velocity is 0, and the speed is equal to the magnitude of the initial horizontal velocity, which is 20 m/s.
(ii) The direction of motion of the ball when the least speed occurs is horizontal.

2. (a) To find the maximum height of the ball, you can use the vertical motion equation: h = ut + (1/2)gt².

The initial vertical velocity u can be found using the given initial velocity and angle:
u = 21 m/s * sin(60°)

Substitute the values into the equation and solve for h. You will find that the maximum height is approximately 16.9 meters.

(b) To find the range of the ball, you can use the horizontal motion equation: s = ut + (1/2)at².

The initial horizontal velocity u can be found using the given initial velocity and angle:
u = 21 m/s * cos(60°)

Since there is no horizontal acceleration, the equation becomes: s = (21 m/s * cos(60°))t

Solve this equation for t and substitute the values to find the range. You will find that the range is approximately 39.0 meters.

(c) To find the speed of the ball at its maximum height, you can use the vertical motion equation: v = u + gt.

The initial vertical velocity u is the same as the value used in part (a), and the acceleration due to gravity is -9.8 m/s².

Substitute the values into the equation and solve for v. You will find that the speed at the maximum height is approximately 10.5 m/s.

3. (a) To find the maximum height of the ball above the landing area, you can use the vertical motion equation: h = ut + (1/2)gt².

The initial vertical velocity u can be found using the given initial velocity and angle:
u = 30 m/s * sin(60°)

Substitute the values into the equation and solve for h. You will find that the maximum height is approximately 38.4 meters.

(b) To find the time taken for the ball to hit the ground, you can use the vertical motion equation: h = ut + (1/2)gt².

The initial vertical velocity u is the same as the value used in part (a), and the acceleration due to gravity is -9.8 m/s².

Substitute the values into the equation and solve for t. You will find that the time taken for the ball to hit the ground is approximately 5.45 seconds.

(c) To calculate the horizontal range of the ball, you can use the horizontal motion equation: s = ut + (1/2)at².

The initial horizontal velocity u can be found using the given initial velocity and angle:
u = 30 m/s * cos(60°)

Since there is no horizontal acceleration, the equation becomes: s = (30 m/s * cos(60°))t

Substitute the values into the equation and solve for s. You will find that the horizontal range is approximately 82 meters.

Please let me know if you need further explanation or assistance with the rest of the questions.