projectile motion question?

Question

A stone thrown off a bridge 20 m above the river has a initial velocity of 12 m /s at an angle of 45 degrees above the horizontal line.a) what is the rane of the stone? b) at what velocity does the stone strike the water?

Answer 1

The Range Equation is:

R = [v^2(sin2theta)] / g.
R = (144)(sin90*) / 9.81
R = 14.68m

For the final velocity, I'd use vf^2 = vi^2 + 2as, where s is the displacement. Since you throw the stone with 12m/s at an angle of 45*, the vertical component going up is 12(sin45). Make this value negative for the initial downward speed of the stone as it travels toward the water.

12(cos45*) = 8.485 so the stone has an initial velocity downward of -8.485 m/s.

Solve for Vf:

Vf^2 = (-8.485^2) + 2(-9.81)(-20m)
Vf = -21.55m/s

Answer 2

a. To solve for the range, we need the horizontal component of the initial velocity and the total time it takes the stone to strike the water.

The horizontal velocity=12cos45
=8.48m/s

To solve for the total time, use the formulas,

v^2-u^2=2as
and
s=ut+1/2at^2

Solve for s first.

Given: v=0, a=9.8m/s^2

we know that u=8.48m/s (this is the vertical component of the velocity, and is equal to the horizontal component because the angle of the initial velocity from the horizontal is 45degrees.)

Substitute known values:

0-8.48^2=2*(-9.8)s
71.91=19.6s
s=71.91/19.6
=3.66m

Solve for t using the formula,

s=ut+1/2at^2

substitute known values:

3.66=0*t+1/2(9.8)t^2
t^2=3.66/4.9
=0.74
t=0.865s

u is the initial velocity from the highest point of the flight of the stone. We know that the vertical component of the velocity at that point equals 0.

Next solve for the time it takes the stone to fall from its highest point to the surface of the river, using the same formula,

s=ut+1/2at^2

s is now the total vertical distance from the highest point to the surface of the river, or, 3.66m+20m=23.66m

substitute known values:

23.66=0*t+1/2*9.8t^2
t^2=23.66/4.9
=4.82
t=2.2s

Thus total time it takes the stone to hit the water equals
0.865+2.2=3.06s

The range =horizontal component of the velocity * total time
=8.48*3.06
=25.9m

b. the vertical component of the velocity of the stone as it hits the water can be solved using the formula,

v^2-u^2=2as
v^2-0=2*9.8*23.66
v^2=463.7
v=21.5m/s

Use phythagorean theorem to compute for the velocity when it hits the water:

final velocity, v^2=21.5^2+8.48^2
final velocity v =(21.5^2+8.48^2)^1/2
=463.7+71.91
=535.6^1/2
=23.1m/s

To get the direction, use arctan8.48/23.1 or arctan0.36. The angle is measured from the vertical axis( use your table to get the corresponding angle of arctan0.36)

Answer 3

well from what i know from my physics class if your finding the final velocety it would be good to find yout final velocety i would recomend you use this equation itll get u your answer. 1/2(m)(v2) = 1/2(m)(Vf)2 x 2(10)times displacement of x.

remember cancel out the variables that you can! and its 10 because that is the normal force. Vf is final velocety so ur gonna solve 4 dat

Answer 4

The damaging sign refers back to the direction of the rigidity, that's downward. by way of convention, damaging speed and damaging acceleration are downward. in case you place g as useful, then a projectile released upwards might have damaging speed, which does no longer make intuitive experience.