Physics-Projectile motion exercises continued

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A cannon ball is fired skywards with an unknown initial vertical velocity and an initial horizontal velocity of 4 m/s as shown on the right. Ignore air resistance
g=10 m/s/s

a) How long does the ball take to reach its maximum height? (use the formula
d =(1/2)gt²
b) What is the initial vertical velocity of the ball?
c) What is the velocity of the ball just before impact at "B"?
d) How far is point "B" from the launch site?

a) How long does the ball take to reach its maximum height? (use the formula
d =(1/2)gt²

200 = (1/2)10t^{2

=> 40 =}t²
=> 6.32s = t

b) What is the initial vertical velocity of the ball?
v =d/t
=>v = 200/6.32
=>v = 31.65 m/s

c) What is the velocity of the ball just before impact at "B"?
The velocity at impact is equal in magnitude but opposite in sign to the initial velocity.
The magnitude of the initial velocity is obtained by adding the vertical and horizontal velocity vectors. Using the Pythagorean Theorem we obtain a value of 31.90 m/s

d) How far is point "B" from the launch site?
To work out the horizontal distance we must use the horizontal velocity. The ball is in the air for 12.64 seconds. So using hte formula d=vt
distance = 4 m/s X 12.64s = 50.56 m.

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Remember that the horizontal and vertical velocities are independent of each other.

To work out the horizontal velocity we use the formula d/t. Now d = 400m and t = ?
To calculate t we use the formula d = 0.5gt^{2

=> 400 = 0.5 X 10 X t²

=> 80 = t²

=> 8.94s = t

So a vertical distance of 400m must be covered in 8.94s. The vertical velocity
must be

v = 400/8.94 = 44.74 m/s.}

^{To calculate
the horizontal velocity again we use the formula v = d/t.

Now d =600 and t = (2 X 8.94) = 17.88s.

So v = d/t

=>v = 600m/17.88s

=>v = 33.56 m/s}

As the cannon ball passes over the peak its vertical velocity has shrunk to zero. It only has horizontal velocity so it passes at a speed of 33.56 m/s

The cannon ball will strike the target at a speed equivalent to its initial speed. We calculate the initial speed by working out the magnitude of the resultant initial velocity vector.
We use Pythagorean Theorem
s² = (hv)² + (vv)² Where s =magnitude of initial velocity vector, hv = initial horizontal velocity and vv = initial vertical velocity.
=> s² = (33.56)² + (44.74)^{2

=> s = 55.93 m/s}

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The ball is thrown horizontally. The initial velocity of the ball is composed of only the horizontal velocity component. Since velocity is calculated by the formula
v = d/t where d =60m and t =?.
To find the time it takes for the ball to travel to the ground we use the formula
d =(1/2)gt² where d =20m, g =10m/s/s
=> t = 2s

So in two seconds the ball travels a horizontal distance of 60m. The horizontal velocity is therefore 30m/s

Projectile motion exercises continued
A cannon ball is fired skywards with an initial vertical velocity of 40 m/s and an initial horizontal velocity of 30 m/s as shown on the right. Ignore air resistance a) What is the initial velocity of the ball? b) How long does the ball take to reach its maximum height? Assume g=10m/s/s/. c) What is the maximum height reached by the ball? d) How far does the ball land from the thrower? e) What is the magnitude of the velocity of the ball at "A"?
A cannon ball is fired skywards with an initial vertical velocity of 70 m/s and an initial horizontal velocity of 240 m/s as shown on the right. Ignore air resistance a) What is the initial velocity of the ball? b) How long does the ball take to reach its maximum height? c) What is the maximum height reached by the ball? e) What is the balls vertical acceleration? f) What is its horizontal acceleration? g) At what point along its path does the ball have minimum speed?
A cannon ball is fired skywards with an initial vertical velocity of 3 m/s and an initial horizontal velocity of 4 m/s as shown on the right. Ignore air resistance a) What is the initial velocity of the ball? b) How long does the ball take to reach its maximum height? c) What is the maximum height reached by the ball? e) What is the balls vertical acceleration? f) What is its horizontal acceleration? g) At what point along its path does the ball have minimum speed?
A cannon ball is fired skywards with an unknown initial vertical velocity and an initial horizontal velocity of 4 m/s as shown on the right. Ignore air resistance g=10 m/s/s a) How long does the ball take to reach its maximum height? (use the formula d =(1/2)gt² b) What is the initial vertical velocity of the ball? c) What is the velocity of the ball just before impact at "B"? d) How far is point "B" from the launch site? Solution
A cannon ball is fired so that it clears a mountain range 400 m above sea level and strikes a target 600 m from the cannon, as shown in the animation on the right. Assume no air resistance and g = 10 m/s/s. Both the cannon and the target are at sea level. How long should the cannon ball take to reach the target? What must the horizontal and vertical velocities be so that the cannon ball just clears the mountain range and strikes the target?. What is the speed of the cannon ball as it strikes the target? What is the speed of the cannon ball as it just passes over the peak of the mountain? Solution
A person stands on a tower and holds a ball so that it is 20 m off the ground. If the ball is thrown 60 m what was the speed with which the ball was thrown? Assume no air resistance. Solution
A person stands on a tower and holds a ball so that it is 30 m off the ground. If the ball is thrown 80 m what was the speed with which the ball was thrown? Assume no air resistance. Try this one on your own. Follow the example above. Answer