What\'s the step by step solution for the two problems using the equations of co

Question

What's the step by step solution for the two problems using the equations of constant acceleration? 2. A projectile is launched from a 16.0 m cliff such that the horizontal component of its initial velocity is 8.2 m/s. The projectile rises 5.5 m above the top of the cliff. (a) What was the initial launch velocity and angle of the projectile. (b) What is the final velocity of the projectile when it hits the ground below the cliff? (no it's not zero) final velocity of the projectle whenit 3. Mrs. Berns wants to try a new form of performance art so she shoots an arrow at a canvas mounted on a wall that is 12.3 m away. The arrow is launched at 35° above the horizontal and leaves the bow 0.75 m above the ground. (a) Will the arrow reach its maximum height before hitting the canvas? Justify your answer. (b) If the arrow's initial velocity is 18.0 m/s, how far above the bottom edge of the canvas will the arrow hit?

Explanation / Answer

(1) Let us consider that the initial velocity is V and it launched at an angle A
So horizontal velocity will be VCosA and vertical velocity = VSinA
Now we know that maximum height reached by the projectile is given by
H = V2Sin2A/2g
VSinA = (2gH)1/2 = (2*9.81*5.5)1/2 = 10.39 m/s ---------(1)
We know the value of VCosA = 8.2 m/s -------------(2)
Now squaring and adding equation 1 and 2
V2 = (10.392 + 8.22)
V = 13.23 m/s this will be the initial speed of the projectile
For launch angle we know that
VCosA = 8.2
therfore angle A = 51.698 degree
Now the velocity when projectile hits the ground
We know that the horizontal velocity remain costant throughout the motion i.e
VCos A = 8.2 m/s
Vertical velocity at ground , we know
V2 = u2 + 2aS
V2 = 2*9.81*(16+5.5)
V =20.53 m/s , vertical velocity at ground.
Net velocity atground
Vg = (20.532 +8.22)1/2 = 22.115 m/s

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