In a scene in an action movie, a stuntman jumps from the top of one building to the top of another building 4.0m away. After a running start, he leaps at an angle of 15 degrees with respect to the flat foor while traveling at a speed of 5.0 m/s. will he make it to the other roof, which is 2.5 m shorter than the building he jumps from? help please, thanks:)!
eth
To determine whether the stuntman will make it to the other roof, we need to analyze the projectile motion of his jump. In this case, we can break down his motion into horizontal and vertical components.
First, let's determine the initial velocity components. The horizontal component (Vx) of the velocity can be calculated using the angle of 15 degrees and the speed of 5.0 m/s:
Vx = velocity * cos(angle)
= 5.0 m/s * cos(15 degrees)
≈ 4.829 m/s
The vertical component (Vy) of the velocity can be calculated using the same angle and speed:
Vy = velocity * sin(angle)
= 5.0 m/s * sin(15 degrees)
≈ 1.289 m/s
Now, we can analyze the vertical motion. We can use the kinematic equation to calculate the time it takes for the stuntman to reach the highest point of his jump. The equation is:
Vy = initial vertical velocity + acceleration * time
1.289 m/s = 0 m/s + (-9.8 m/s^2) * t
Solving for time (t), we get:
t = 1.289 m/s / 9.8 m/s^2
≈ 0.132 s
Next, we can use this time to find the maximum height (H) reached by the stuntman using the following equation:
H = Vy * t + 0.5 * acceleration * t^2
= 1.289 m/s * 0.132 s + 0.5 * (-9.8 m/s^2) * (0.132 s)^2
≈ 0.086 m
Now, we can analyze the horizontal motion. The horizontal distance (d) traveled can be calculated using the horizontal velocity (Vx) and the time of flight (2t) since the total time of flight is twice the time it took to reach maximum height:
d = Vx * time of flight
= 4.829 m/s * 2 * 0.132 s
≈ 1.278 m
Finally, we consider the difference in height between the two buildings. If the second building is 2.5 m shorter than the first building, the stuntman needs to clear a horizontal distance of 4.0 m and a vertical distance of 2.5 m.
Since the stuntman travels a horizontal distance of 1.278 m and the second building is only 4.0 m away, he will make it to the other roof horizontally.
However, since the maximum height he can reach is approximately 0.086 m and the second building is 2.5 m shorter, he will not clear the vertical distance. Therefore, he will not make it to the other roof.
Note: This analysis assumes a simplified model of projectile motion and does not account for factors such as air resistance.
het is me
Calculate how far he travels horizontally before falling 2.5 meters. If the result is 4.0 m or more, he survives the jump.
The horizintal distance travelled equals the horizontal jumping velocity component (5.0 cos 15 m/s) times the time it takes to fall 2.5 m. To calculate that time, you need to take into consideration the initial vertical velocity component of 5.0 sin 15 = 1.29 m/s.
With hints, you should be able to proceed. Show your work if futher held is needed.