Please can anyone help me with my Projectiles work? I would be grateful if any workings out could be shown.
Cannon balls fired from a sail ship travel a horizontal distance of 192m in 1.6s before striking the sea. Ignore air resistance.
1) Calculate the horizontal velocity of the cannon balls.
2) Determine the height of the cannon above the sea.
3) Calculate the resultant velocity of the cannon balls as they strike the sea.
4) The horizontal velocity is now increased. Explain the effect, if any, that this would have on the time of the flight, i.e. the time taken for the cannon balls to strike the sea.
5) Discuss the effect that air resistance would have on the cannon ball's motion.
Any help would be appreciated.
Cannon balls fired from a sail ship travel a horizontal distance of 192m in 1.6s before striking the sea. Ignore air resistance.
1) Calculate the horizontal velocity of the cannon balls.
2) Determine the height of the cannon above the sea.
3) Calculate the resultant velocity of the cannon balls as they strike the sea.
4) The horizontal velocity is now increased. Explain the effect, if any, that this would have on the time of the flight, i.e. the time taken for the cannon balls to strike the sea.
5) Discuss the effect that air resistance would have on the cannon ball's motion.
Any help would be appreciated.
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Hello
1) horizontal velocity = s/t = 192/1.6 = 120 m/s
2) 1.6 seconds is the time needed to go up to the top height and down again
--> the time needed to go up is 0.8 seconds.
and the time needed to reach the maximum height h is v0v/g with vov = initial vertical velocity
vov = gt = 9.81*0.8 = 7.848 m/s.
The top height reached with this initial vertical velocity is
hmax = v0v^2/(2g) = 7.848^2/(2*9.81)
hmax = 3.1392 m
3) the initial velocity v0 = √(vh^2 + v0v^2) = √(120^2 + 7.848^2)
v0 = 120.256 m/s
and this is also the velocity which which the ball hits the ground, because the trajectory is symmetric.
4) If the horizontal velocity is increased, this would not affect the time of flight. What matters for the time of flight is the vertical initial velocity.
5) the horizontal distance reached, and also the maximum height, would be smaller.
Regards
1) horizontal velocity = s/t = 192/1.6 = 120 m/s
2) 1.6 seconds is the time needed to go up to the top height and down again
--> the time needed to go up is 0.8 seconds.
and the time needed to reach the maximum height h is v0v/g with vov = initial vertical velocity
vov = gt = 9.81*0.8 = 7.848 m/s.
The top height reached with this initial vertical velocity is
hmax = v0v^2/(2g) = 7.848^2/(2*9.81)
hmax = 3.1392 m
3) the initial velocity v0 = √(vh^2 + v0v^2) = √(120^2 + 7.848^2)
v0 = 120.256 m/s
and this is also the velocity which which the ball hits the ground, because the trajectory is symmetric.
4) If the horizontal velocity is increased, this would not affect the time of flight. What matters for the time of flight is the vertical initial velocity.
5) the horizontal distance reached, and also the maximum height, would be smaller.
Regards