s= vt+1/2gt^2
v2 = v02 + 2gs
0 = 352 - 2*9.8*s (taking down as the positive direction)
62.5 meters
63
Initial upward speed = 7.61 m/sFinal upward speed (at the point of maximum height) = 0Time to reach maximum height = (7.61) / (9.8) = 0.77653 secondAverage speed during that time = 1/2 ( 7.61 + 0) = 3.805 m/sHeight = 3.805 x 0.77653 = 2.9547 meters (rounded) = about 9.7 feetDoesn't seem like much of a height for a strong toss; but the math looks OK.
If the initial velocity is 50 meters per second and the launch angle is 15 degrees what is the maximum height? Explain.
applying the equation: (VxV) -(UXU) = 2aS on reaching the maximum height the ball stops so the final velocity(V) becomes =0. initial velocity as given is (U)=10 m/s. ------------------------------------------------------------------------------------2 acceleration is here gravity in downward direction so (a)=-9.81 m/s so applying the above stated formula the height reached by the ball(S)=5.0968 m (approx)
Making the improbable assumption that the jumper experiences no air resistance, he will jump 3.97 metres, and reach a height of 0.72 metres.
'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
The time taken by the ball to reach the maximum height is 1 second. The maximum height reached by the ball is 36 meters.
When it's at its maximum height its speed will be zero.
amplitude
height=acceletation(t^2) + velocity(t) + initial height take (T final - T initial) /2 and place it in for time and there you go
Ignoring air resistance, I get this formula:Maximum height of a vertically-launched object = 1.5 square of initial speed/GI could be wrong. In that case, the unused portion of my fee will be cheerfully refunded.
265
Height reached = 3.7 metres.The mass of the ball is not really relevant.
An archer uses a force of 60.0N to draw back the string of his bow through a distance of 0.330m. He then fires a 300g arroe straight up into the air. What is the maximum speed of the arrow at the instane it leaves the bow? What is the maximum height reached by the arrow in its flight into the air?
19, when the tree went from 19 to 20, it's height doubled to maximum.9 Years.It would have doubled it's height on it's 10th year
By releasing the ball at or just before you have reached you maximum jumping height for the shot.
algebra 2 right? i hated that unit man i forgot everything we learned in that class
initial velocity of the kick = 28.06 m/s