velocity of 10 m/s at an angle of 37°. height was 2.0 mass is 7.27 kg. Find the distance traveled by the shot in this effort.

Answer 1

I am assuming this is in regard to the sport of shot put, and the height of the thrower is such that the shot will be 2 meters off of the ground at release.

I am also assuming that the angle of 37^o refers to the angle up from the ground to the initial shot trajectory. Therefore the shot has an initial upward velocity of (10 m/s)*(sin(37)), and it has an initial horizontal velocity of (10 m/s)*(cos(37)).

Neglecting air resistance, we can use the equation of vertical position (Y) for an object with some initial position (Y_o), velocity (Vy_o) and acceleration due to gravity (9.806 m/s² downward).

Since the shot is starting from 2 m off of the ground, traveling upwards, then back down, it must pass the 2 m high point before returning to the ground. The time that it takes to go up then back down to the 2 m high point can be found from:

Y = Y_o + (Vy_o)*(t) -(1/2)*(9.806 m/s²)*(t)²

Substituting known quantities gives:

2 m = 2 m + [(10 m/s)*sin(37)]*(t) - (1/2)*(9.806 m/s²)*(t)²

[(10 m/s)*sin(37)]*(t) = (1/2)*(9.806 m/s²)*(t)²

[(10 m/s)*sin(37)] = (1/2)*(9.806 m/s²)*(t)

t = [(10 m/s)*sin(37)] / [(1/2)*(9.806 m/s²)]

t = 1.227 s

So it takes 1.227 seconds for the shot to rise up in its trajectory and fall back to the starting height of 2 m off of the ground.

The remainder of the time that the shot spends in the air is the time it takes to fall from 2 m high. Using the equation Vy = Vy_o - (g)(t), with Vy_o = [(10 m/s)*sin(37)] , and g=9.806 m/s², we can find the vertical velocity of the shot as it passes the 2 m high mark. Vy = -6.01 m/s (which is approximately the opposite of the initial vertical velocity upwards). Using an initial vertical velocity of -6.01 m./s and an acceleration of -9.806 m/s² with an initial height of 2 m, we can form the equation:

Y = Y_o + (Vy_o)*(t) -(1/2)*(9.806 m/s²)*(t)²

0 = 2 m +(-6.01 m/s)*(t) - (4.903 m/s²)*(t)²

rearranging into quadratic form:

(t)² + (1.227)*(t) - (0.4079) = 0

[solve by quadratic equation solution]

Here the negative value of (t) can not be used, so t = 0.272 s.

So it takes an additional 0.272 s until the shot hits the ground. Adding the 1.227 s it took for the shot to go from 2 m, into the air, and back down to 2 m, gives a total of 1.5 s.

In 1.5 seconds, the shot goes a horizontal distance of [(10 m/s)*cos(37)] *(1.5 s) = 12 m.

Answer 2

Here's an outline of the solution to the problem:

1. Find the x- and y- components of velocity.
2. Find out how long the ball is in the air by using the y-component of the velocity.
3. Find out how far the ball travelled by using the x-component of the velocity and the time that the ball spent in the air.

Step 1. We have the velocity in the form of a magnitude and a direction, but we need it in the form (x,y). We recall from basic trigonometry that the x-component of velocity will then be 10 * cos 37° = 7.99 m/s and the y-component of velocity will be 10 * sin 37° = 6.02 m/s.

Step 2. The displacement of the ball in the y direction over the whole time that the ball is in the air is equal to -2 because it starts at a height of 2 meters and ends at a height of 0 meters. We also know the y-component of velocity, and do not know the amount of time the ball spends in the air, so we know Δy = -2 m, v_y = 6.02 m/s. Therefore, we should use the kinematic equation Δy = v_yt + (1/2)at^2, where a = 9.81 m/s^2 (we're on Earth after all). Then you can just use the quadratic formula to find the value of t. (I omit this part of the solution so that you can do some of this yourself)

Step 3. Since the ball is in the air for t seconds, we know that in the x-direction, it will have a displacement Δx = v_xt = 7.99 t meters, as there is no acceleration in the x direction. Thus the distance that the ball travels is 7.99 t meters.