Q: A ball is thrown with a velocity of 40 meters per second at an angle above the horizontal. If the ball attains its maximum range what was the value of the angle?

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If the initial velocity is 50 meters per second and the launch angle is 15 degrees what is the maximum height? Explain.

The speed of an object falling from a great height is measured in meters per second per second until it reaches terminal velocity (maximum downward speed).

'Maximum height' means the exact point at which the velocity changes from upward to downward. At that exact point, the magnitude of the velocity is zero. It doesn't matter what the velocity was when it left your hand. That number determines the maximum height, but the velocity at that height is always zero. --------------------------------------------------------- Thus using the formula: (vf)e2 = (vi)e2+2*a*d vf = final velocity = 0 m/s vi = initial velocity = 10 m/s a = acceleration = gravity = - 9.81 m/s/s d = displacement (distance) = ? e is designating that the next figure is an exponent in the formula So the formula is: (0)e2 = (10)e2 + (2 * -9.81 * d) 0 = 100 + -19.62d adding 19.62d to both sides of the equation 19.62d = 100 dividing by 19.62 d = ~ 5.097 meters

Acceleration of the arrow is -3m/s2A = (velocity minus initial velocity) / time

Yes.

Related questions

At the maximum height of projectile motion, the vertical component of velocity is zero while the horizontal component of velocity remains constant. Therefore, the total velocity of the projectile at the maximum height is equal to the magnitude of its horizontal component of velocity.

There are two different ways of looking at this. The maximum length from end to end from the telecommunications closet to the wor karea device is 100 meters. The horizontal cable itself is usually a maximum of 90 meters, with 10 meters maximum from the wall plate to the device in the work area.

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The acceleration at instantaneous maximum velocity is zero, as the velocity is not changing at that moment.

The horizontal component of a projectile's velocity doesn't change, until the projectile hits somethingor falls to the ground.The vertical component of a projectile's velocity becomes [9.8 meters per second downward] greatereach second. At the maximum height of its trajectory, the projectile's velocity is zero. That's the pointwhere the velocity transitions from upward to downward.

Using the projectile motion equations and given the initial velocity and angle, we can calculate the time the shell is in the air. Then, we can find the horizontal range by multiplying the time of flight by the horizontal component of the initial velocity. The horizontal range in this case is about 1056 meters.

90 meters

90 Meters+ 10 meters for running cables so 100 meters is the total.

There is not enough information to answer the question. The initial velocity of the car is not given. Also, the "it finally" at the end of the question does not make sense.

To find the horizontal distance traveled by the soccer ball, you can use the equation: horizontal distance = horizontal velocity x time. The horizontal velocity is given by the formula Vx = V0 cosθ, where V0 is the initial velocity and θ is the angle of projection. Substituting the given values: Vx = 10.0 m/s * cos(30°) = 8.66 m/s. Then, the horizontal distance = 8.66 m/s * 3.2 s = 27.71 meters.

After a second, the ball will still have a horizontal velocity of 8 meters per second. It will also have a vertical velocity of 9.8 meters per second (Earth's acceleration is about 9.8 meters per square second). The combined speed (using the Law of Pythagoras) is about 12.65 meters per second.

If the initial velocity is 50 meters per second and the launch angle is 15 degrees what is the maximum height? Explain.