A ball is thrown with a horizontal velocity of 2 m/s from a vertical height of 50 m above ground. neglect air resistance and determine:
the horizontal displacement traveled by the ball. can someone please explain this to me!
One needs to find how long it is in the air, the initial height determines that.
h=1/2 g t^2
Then, knowing the time in air, the horizontal distance is
d=vhorizontal*time
Thanks!! Could you explain this one to me?
object is projected with a horizontal velocity of 30 m/s. if it hits the ground 5 seconds later, calculate:
The height from which the particle was projected.
h=1/2 g t^2
memorize that equation.
calculate height.
thanks!
To determine the horizontal displacement traveled by the ball, we need to consider its horizontal velocity and the time it takes to reach the ground.
The horizontal velocity remains constant throughout the motion, as there is no horizontal force acting on the ball. In this case, the horizontal velocity is given as 2 m/s.
To find the time it takes for the ball to reach the ground, we can use the vertical motion equation:
h = (gt²) / 2,
where h is the initial vertical height (50 m in this case), g is the acceleration due to gravity (9.8 m/s²), and t is the time.
Rearranging the equation to solve for time, we have:
t = √(2h / g).
Substituting the given values into the equation, we get:
t = √(2 * 50 m / 9.8 m/s²) ≈ √(10.2) ≈ 3.19 s.
Now that we have the time it takes for the ball to reach the ground, we can find the horizontal displacement using the formula:
d = v * t,
where d is the horizontal displacement and v is the horizontal velocity.
Plugging in the values, we get:
d = 2 m/s * 3.19 s ≈ 6.38 m.
Therefore, the horizontal displacement traveled by the ball is approximately 6.38 meters.