I'm sorry for the repost I just cant seem to find the answer for this. I did this in A1 and got it all wrong and was given the answer sheet and my teacher told me to look at that which is below, but I didnt understand and shes busy. I understand projectiles except for all these questions in this worksheet. For some reason i dotn understand this worksheet. Can someone who knows physics please show working out and equations used on all these questions. I lost my working out but i have been trying for the second year and still cant get my head around these questions though i always get projectiles right in other questions/worksheets. Please help.
Projectiles Exam Questions
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
Oh, I just found Bob Pursleys answer, I'll upload my working out for it soon so you can see where I am going wrong.
http://www.learncoombedean.com/pluginfile.php/2287/mod_resource/content/0/M1_07_Projectiles.pdf
1. Vo = 20m/s[53o].
Xo = 20*Cos53 = 12.04 m/s.
Yo = 20*sin53 = 15.97 m/s.
a. Y = Yo + g*Tr = 0.
15.97 + (-9.8)Tr = 0,
Tr = 1.63 s. = Rise time.
Tf = Tr = 1.63 s. = Fall time.
T = Tr + Tf = 3.26 s.
b. d = Xo * T = 39.2 m.
c. I. V = Xo = 12 m/s(Y component = o).
ii. Direction: Hor.
2. Vo = 21m/s[60o].
Xo = 21*Cos60 = 10.5 m/s.
Yo = 21*sin60 = 18.19 m/s.
a. Y^2 = Yo^2 + 2g*h = 0.
18.19^2 + (-19.6)h = 0,
h = 16.9 m.
b. Y = Yo + g*Tr = 0.
18.19i + (-9.8)Tr = 0,
Tr = 1.86 s. = Rise time.
Tf = Tr = 1.86 s. = Fall time.
Range = Xo*(Tr+Tf) = 39 m.
c. V = Xo = 10.5 m/s.
4a. Vo = 10m/s[70o].
Xo = 10*Cos70 = 3.42 m/s.
Yo = 10*sin70 = 9.40 m/s.
Y = Yo + g*Tr = 0..
9.40 + (-9.8)Tr = o,
Tr = 0.959 s. = Rise time.
Tf = Tr = 0.959 s. = Fall time.
T = Tr + Tf = 0.959 + 0.959 = 1.92 s. = Time of flight.
Range = Xo * T = 3.42 * 1.92 = 6.56 m.
Q4.
a) resolving vertically s=0 u=10sin70 a=-9.8 t=?
using s=ut+1/2at^2
time of flight using the simple model is 1.92s
b) we are given different parameters therefore we must remodel and recalculate the tie of flight
resolving vertically, s=-2 u=10sin70 a=-9.8 t=?
using s=ut+1/2at^2
t=2.11s
therefore, when resolving horizontally, using s=ut
s= 10cos70 x 2.11
s= 7.22m (3sf)