Physics projectile motion?

Question

In the Olympic shotput event, an athlete throws the shot with an initial speed of 12 m/s at a 40.0^\circ angle from the horizontal. The shot leaves her hand at a height of 1.8 m above the ground.

How far does the shot travel?

Answer 1

This is a straight forward projectile motion problem.

You are given the initial values for velocity and launch height, plus the assumed gravitational acceleration of -9.81 m/s^2.

Break the initial velocity up into its vertical and horizontal components. Find the time it takes for the ball to hit the ground. Use this time to find how far (horizontally) the ball travels.

Use the following formulas,
d_y = v_i_y * t + 1/2 * a * t^2
d_x = v_x * t

v_x = v * cos (theta)
v_y = v * sin (theta)
where v is the initial launch velocity.

where d is the distance traveled, v is velocity, a is the acceleration, t is time, and theta is the launch angle. The _ characters indicate subscripts...._x and _y specify in the x and in the y direction, and _i is initial.

Answer 2

There are two components to this motion, x and y

y = y0 + vy0*t - (1/2)*9.8*t^2

x = x0 + vx0*t

You are looking for the horizontal distance where the shot lands, I assume, which is "x - x0"

From above, x - x0 = vx0*t

vx0 and vy0 are the components of the initial velocity in the x and y directions respectively. Since you know the angle, you can use the total initial velocity, v0 = 12 m/s to find both vx0 and vy0.

The v0 is the 'hypotenuse', vx0 the 'adjacent' and vy0 the 'opposite'. So,

sin 40 = vy0 / v0 ----> vy0 = 7.71 m/s
cos 40 = vx0 / v0 ----> vx0 = 9.19 m/s

Start substituting:

y = 1.8 + 7.71*t - (1/2)*9.8*t^2

x - x0 = 9.19*t

The trick is to find the time "t" when the shot has reached its final position and then solve for "x - x0". you do know that when the ball lands, y = 0, so equate the first equation to zero and solve for "t"

0 = 1.8 + 7.71*t - (1/2)*9.8*t^2

Use the quadratic formula and solve for t
t = 1.78 seconds

Thus, x - x0 = 9.19*t = 9.19*1.78 = 16.36 meters.

Answer 3

As long the problem is relative to the ground height of throwing is not a factor. Just increase the height considerably and imagine yourself. Just because you are throwing from top of building does not make an object travel longer distance all other things being same.

This is a trajectory problem and the range of parabola figure will give the distance that shot travels.

Range = 2*(hor.comp of initial vel)*(ver.comp of init.vel) / g

= 2 * (12*cos40)*(12*sin40) / 9.8

calculated value will be in meters. hope it helps