Question

A ball is thrown upward with an initial velocity of 30 m/s. How high will it rise? What will be the velocity of the ball when it returns to its original position?

Answer 1

No there isn't

Answer 2

As the ball moves upward, the vertical velocity of an object decreases 10 m/s each second.

Answer 3

As the ball moves downward, the vertical velocity of an object increases 10 m/s each second. The path is symmetrical. The time up equals the time down. The final vertical velocity, after falling back down is equal to the initial vertical velocity

Answer 4

first we have to calculate the rise time...... as we know Vi = 30 m /s Vf = 0 we know that a = \[\Delta v /t\]

Answer 5

a = g = 9.8 m/sec^2 so by finding \[\Delta v\] we have V f - V i = 0 - 30m/s= -30m/s
and when ball thrown upward value of g is negative..... so for time we have the equation \[t = \Delta v /g = -30 / - 10 = 3 s\]

Answer 6

Now find the height of the ball.. with the 2nd equation of motion. that is....
\[S = Vi t + 0.5 a t ^{2}\]

Answer 7

as ball moves upward... then s = h... . And in the upward motion a = g = - 10 m / s^2
h = ( 30)(3) + (0.5)(-10)\[(3)^{2}\]

Answer 8

height = 90 - 45 = 45 meters

Answer 9

And at the orignal point, the velocity of the ball will be same as it was before.

Answer 10

the initial speed of the ball is 30m/s.its 30 meter per second..which is a good velocity.it will form a parabola in the field.and it will rise till v-max which means the maximum velocity of the parabola;the traveling path captured by the ball,after v-max..it will reduce its path of going completing the 2nd half of the parabola.the final speed is zero.so apply the law needed for this incident..if the ball is thrown upward straight,,it will come back to ur hand after reaching vmax,.depends on how you threw it..what angles in vaccum or in normal state.

Answer 11

@IhteshamMalik thanks!

Answer 12

Its my pleasure.