A projectile is shot upward at a 60° angle with the ground at 65 m/s. How far in meters has the projectile gone horizontally after 4.0 seconds?
the horizontal speed is a constant 65 cos60° = 32.5 m/s
so, what do you think?
To find the horizontal distance covered by the projectile after 4.0 seconds, we can use the formula:
Distance = (horizontal velocity) * (time)
The horizontal velocity of the projectile remains constant throughout its flight. To find the horizontal velocity, we can use the following formula:
Horizontal velocity = (initial velocity) * (cos(angle))
In this case, the initial velocity is 65 m/s, and the angle is 60°. So we can substitute these values into the formula:
Horizontal velocity = 65 m/s * cos(60°)
Simplifying further:
Horizontal velocity = 65 m/s * 0.5
Horizontal velocity = 32.5 m/s
Now, we can substitute the value of time (4.0 seconds) into the first formula to find the horizontal distance covered by the projectile:
Distance = 32.5 m/s * 4.0 s
Distance = 130 meters
Therefore, the projectile has traveled a horizontal distance of 130 meters after 4.0 seconds.
To find the horizontal distance covered by the projectile after 4.0 seconds, we can use the basic kinematic equations of motion.
In this case, we are given the initial velocity of the projectile, which is 65 m/s, and the time, which is 4.0 seconds.
To find the horizontal distance, we need to determine the horizontal component of the velocity. The horizontal and vertical components of the initial velocity can be calculated as follows:
Horizontal component:
Vx = V * cos(θ)
= 65 m/s * cos(60°)
= 65 m/s * 0.5
= 32.5 m/s
Vertical component:
Vy = V * sin(θ)
= 65 m/s * sin(60°)
= 65 m/s * 0.866
= 56.29 m/s
Since we are only interested in the horizontal distance traveled, we can ignore the vertical component. The horizontal distance covered can be calculated using the equation:
Distance = Vx * time
= 32.5 m/s * 4.0 s
= 130 meters
Therefore, after 4.0 seconds, the projectile has covered a horizontal distance of 130 meters.