Projectile Motion

Projectile Motion
	While solving problems in projectile motion, I expect students to solve these problems by drawing a vertical line on their paper and write Vertical Motion on the first column and Horizontal Motion on the second column because in projectile motion, the Vertical Motion is INDEPENDENT of the Horizontal Motion.
	VERTICAL MOTION	HORIZONTAL MOTION

1.	This is a problem that we consider as horizontal projection.
	Here you're given all the information for the vertical motion. To find the height of the table there is no need for the initial speed.	Assuming horizontal projection, we would need 0.24 ms^-1to calculate where the ball would strike horizontally on the table.

	Now have you drawn two columns to start reasoning logically. For vertical motion, you can write the equation of motion as d = 1/2 gt² There is no initial velocity in the vertical direction (and therefore u = 0)

	Substituting the values you'll have d = 1/2 (9.8)(0.3)² = 0.44 m (2 significant figures)

2.	This is another "horizontal projection" problem.

	For vertical motion, you can write the equation of motion as d = 1/2 gt² There is no initial velocity in the vertical direction (and therefore u = 0) Substituting the values you'll have 1.0 = 1/2 (9.8)(t)² ⇒ t = 0.452 s (3 significant figures)	Using horizontal motion now, you can similarly write the equation of motion as d = ut (since 1/2at² = 0) There is no acceleration in the horizontal direction (and therefore a = 0) Substituting the values you'll have 1.0 = u (0.452) ⇒ u = 2.2 m/s (2 significant figures) and the correct choice is therefore (B)

3.	This is a classic problem of Projection at an Angle

	a) To Find H:	c) To Find R:

	u = 30sin60⁰ = 26 m/s a = -g = -9.81 m/s² v = 0 (at maximum height) d = H = ? Using the equation v² = u² + 2ad, we have 0 = (26)² + 2(-9.81)H ⇒ H = 34.5 m	u = 30cos60⁰ = 15 m/s a = 0 t = 5.3 s d = R = ? Using the equation d= ut + 1/2at², we have R = 15(5.3) ⇒ R = 79.5 m

	b) To Find t:

	u = 30sin60⁰ = 26 m/s a = -g = -9.81 m/s² d = 0 (displacement is zero when projectile goes up & comes down in the vertical direction) t = ? Using the equation d= ut + 1/2at², we have 0 = (26)t + 1/2(-9.81)t² Dividing both sides by t 0 = 26 - 4.91t ⇒ t = 5.3 s

9.	It's really another sad day for the Coyote because he never gives up despite the immense support for his ideas with gizmos supplied by Acme. The first thing you must recognize with this problem is one involving the equations of motion and 'horizontal projection'.

	a) You can find the time Coyote takes to travel the distance of 70 m starting from rest and having an acceleration of 1.5 ms^-2. Using the equation d = 1/2 at² (since u = 0) You'll find t = 9.7 s. In this time, the road runner has to cover the same 70 m. Therefore the mimimum speed for the road runner must be v = s/t (since a = 0) = 70/9.7 = 7.2 ms^-1 (2 s.f.) to reach before the Coyote. b)To solve this part of the problem you must first know the time Coyote is going to be in air. Considering only the VERTICAL motion, you'll find d = 1/2 g t² 100 = 1/2 (9.8) t² Giving you t = 4.5 s. (2 s.f.)	In this time, you can find the HORIZONTAL distance travelled on the canyon by Coyote. To do this you must also find the velocity of the Coyote just as he leaves the cliff. Using the equation v = at (since u = 0), you'll have v = (1.5)(9.7) = 14.6 ms^-1 This is the initial velocity of Coyote just before leaving the cliff. Using the equation d = ut + 1/2 at² for the HORIZONTAL motion, you'll have d = (14.6)(4.5) + 1/2 (1.5)(4.5)² = 81 m This is the distance where the Coyote lands on the canyon.

	c) The component of the VERTICAL velocity would be v = gt (since u = 0) in the vertical directon. = (9.8)(4.5) = 44.1 ms^-1	c) The component of the HORIZONTAL velocity would be v = u + at = (14.6) + (1.5)(4.5) = 21.4 ms^-1 (3 s.f.)
	After all this, it is indeed a pity that the road runner has turned off!

10.	As usual, first find the velocity of the man as he reaches the edge of the roof. Then resolve the velocity into the VERTICAL and HORIZONTAL components. The vertical accceleration is g = 9.8 ms^-2 and the horizontal acceleration is a = 0.

	a) To find the time and velocity just as he reaches the edge of the roof: x = (v_i)(t) + 0.5(a)(t²) 8= 0.5(5)t² giving you, the time t= 1.79 ~ 1.8 seconds	c) Finally to find the distance between the house and the point where he lands in the snow,consider the HORIZONTAL motion. Remember all along you were only considering the vertical motion. Using the same equation you had in the beginning, you'll have x = v_it + 0.5at² = (7.2)(0.68) + 0.5 (0)(0.68)² = 4.9 + 0 = 4.9 m

	Using this time you can find the speed of the man when he reaches the edge of the roof. Using v_f = v_i + at = 0 + (5)(1.8) = 9 ms^-1 This velocity acts as the initial velocity from the edge of the roof. The vertical component would be 9sin37 = 5.4 ms^-1 and the horizontal component would be 9cos37 = 7.2 ms^-1. Considering the vertical motion, v_i = 5.4 ms^-1 d = 6 m a = 9.8 ms² We have a choice here. Either you can use the same equation you used, but it would result in a quadratic equation in 't'. The better idea would be to use the equation v_f² = v_i² + 2a.d and find v_f. This is easier even if it involves squaring. Substituting you'll have v_f² = (5.4)² + 2(9.8)(6) giving you v_f = 12.1 ms^-1 downward. This is the vertical component of velocity when he reaches the snow bank.

	b) Now it is easy to find the time of motion. Using the equation v_f = v_i + at and making 't' the subject of this equation, you'll have t = v_f - v_i/a = 12.1 - 5.4/9.8 = 0.68 s


	For vertical motion, you can write the equation of motion as d = 1/2 gt² There is no initial velocity in the vertical direction (and therefore u = 0)	Using horizontal motion now, you can similarly write the equation of motion as d = ut (since 1/2at² = 0) There is no acceleration in the horizontal direction (and therefore a = 0)

	Substituting the values you'll have 1.0 = 1/2 (9.8)(t)² ⇒ t = 0.452 s (3 significant figures)	Substituting the values you'll have 1.0 = u (0.452) ⇒ u = 2.2 m/s (2 significant figures)