<h2>
Component of the velocity of the ball in the horizontal direction just before the ball hits the ground = 7.31 m/s</h2>
Explanation:
In horizontal direction there is acceleration or deceleration for a ball tossed upward at an initial angle of 43° off the ground.
So the horizontal component of velocity always remains the same.
Horizontal component of velocity is the cosine component of velocity.
Initial velocity, u = 10 m/s
Angle, θ = 43°
Horizontal component of velocity = u cosθ
Horizontal component of velocity = 10 cos43
Horizontal component of velocity = 7.31 m/s
Since the horizontal velocity is unaffected, we have
Component of the velocity of the ball in the horizontal direction just before the ball hits the ground = 7.31 m/s
Displacement = final - initial
The formula most closely resembling that is delta = xf - xi
The period of the wave is the reciprocal of its frequency.
1 / (5 per second) = 0.2 second .
The wavelength is irrelevant to the period. But since you
gave it to us, we can also calculate the speed of the wave.
Wave speed = (frequency) x (wavelength)
= (5 per second) x (1cm) = 5 cm per second
Answer:
2.8125 meters
Explanation:
I will assume 7.5 m/s
When the Kinetic Energy is entirely converted to Potential energy the max height will have been reached
KE = PE
1/2 m v^2 = mgh divide by m
1/2 v^2 = gh looking for h so divide both sides by
1/2 v^2 / g = h sub in the values using g = 10 m/s^2
1/2 ( 7.5 )^2 / 10 = 2.8125 m
