A ball of mass 500g is kicked into the air at angle of 45degree. It reaches a height of 12m. What was it's initial velocity?
2 g y = (v sin45)^2
solve for v
v = [sqrt(2gy)]/sin45
mgh =1/2mv"2
v"2= 2gh
v"2=2x10,x12
v=15.5
21m/s
Excellent. know I need the answer of Kinetic energy and potential energy
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To find the initial velocity of the ball, we can use the equations of motion for projectile motion. Let's assume the ball was launched with an initial velocity "v0" at an angle of 45 degrees.
The vertical motion of the ball can be described by the equation:
Δy = v0 * sinθ * t - (1/2) * g * t^2,
where Δy is the vertical displacement (in this case, the height of 12m), v0 is the initial velocity, θ is the launch angle (45 degrees), t is the time of flight, and g is the acceleration due to gravity (approximated as 9.8 m/s^2).
We know that at the maximum height, the vertical displacement, Δy, is 12m. Taking the initial launch angle as 45 degrees, we can use the equation to find the time of flight, t.
12 = v0 * sin45 * t - (1/2) * 9.8 * t^2.
Now we need to solve this quadratic equation to find the time of flight, t.
0.5 * 9.8 * t^2 - v0 * 0.707 * t + 12 = 0.
Solving this equation will give us the time of flight, t.
Once we have the time of flight, we can substitute it back into the original equation to find the initial velocity, v0.
v0 = 12 / (t * sin45).
By calculating the value of t using the quadratic equation and substituting that value into the equation for v0, we can find the initial velocity of the ball.