How Do You Calculate Projectile Motion for a Stone Thrown from a Tower?

[rachael]

A stone is prjected with a velocity of 100m/s at an elevation of 30 degrees form a tower 150m high. Find:
a. the time of flight
ans:12.(4) secs
b. the horizontal distance from the tower at which the stone strikes the ground
ans: 1.0(8) x 10 3
c. the magnitude and the direction of the velocity of the stone striking the ground.
Ans: 114m/s at 40(5) degrees below horizontal


THANKS

 

[finchie_88]

1. $$y = 30tsin30 - 4.9t^2 + 150$$
solve that, when y = 0.
2. $$x = 30tcos30$$ use the value for t that you just worked out.
3. $$v_y = 30sin30 - 9.8t$$
$$v_x = 30cos30$$
Solve these two, using the value of t you got in part 1, and then use pythagoras

 

[kyle8921]

FOR YOUR QUESTION! I am happy to provide a response to your inquiry about projectile motion. First, let's define projectile motion - it is the motion of an object through the air, under the influence of gravity, after being launched at an angle. In this case, the stone is being launched from a tower at an angle of 30 degrees and with a velocity of 100m/s.

To find the time of flight, we can use the equation t = 2v*sin(theta)/g, where t is the time of flight, v is the initial velocity, theta is the angle of elevation, and g is the acceleration due to gravity. Plugging in the given values, we get t = 2*100*sin(30)/9.8 = 12.4 seconds. Therefore, the time of flight is 12.4 seconds.

To find the horizontal distance from the tower at which the stone strikes the ground, we can use the equation x = v*cos(theta)*t, where x is the horizontal distance, v is the initial velocity, theta is the angle of elevation, and t is the time of flight. Plugging in the given values, we get x = 100*cos(30)*12.4 = 1.08*10^3 meters. Therefore, the horizontal distance from the tower at which the stone strikes the ground is 1.08*10^3 meters.

To find the magnitude and direction of the velocity of the stone striking the ground, we can use the Pythagorean theorem and trigonometric functions. The magnitude of the velocity can be found using the equation v = sqrt(vx^2 + vy^2), where vx is the horizontal component of the velocity and vy is the vertical component of the velocity. From our previous calculations, we know that vx = 100*cos(30) = 86.6 m/s and vy = 100*sin(30) = 50 m/s. Plugging these values into the equation, we get v = sqrt(86.6^2 + 50^2) = 114 m/s. Therefore, the magnitude of the velocity is 114 m/s.

To find the direction of the velocity, we can use the inverse tangent function, tan^-1(vy/vx). Plugging in the values, we get tan^-1(50/86.6) = 40.5 degrees. However, this angle is measured