# What will the ball's velocity be when it returns to its starting point if a ball is thrown vertically upward with an initial velocity of +7.5 m/s?

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As the ball travels upward its velocity will constantly decrease due to the downward effect of gravity acting on the ball. Eventually the velocity of the ball will reach a value of zero at the highest point of the ball's flight. Then the ball will start falling back to earth, gaining speed due to the influence of gravity on the ball. When the ball is moving upward, earth's acceleration due to gravity, 9.8 m/s/s is acting against the ball, while on the return to earth, that same 9.8 m/s/s is causing the velocity of the ball to increase.

Because the ball is moving in a parabolic motion, ignoring any effects due to air resistance and wind, the velocity of the ball when it returns to the starting point has the same magnitude, 7.5 m/s, but because velocity is a vector, with both magnitude and direction, the new velocity is -7.5 m/s.

Because you defined up as positive initially (+7.5 m/s), that means that velocity in the opposite direction (down in this case) must be negative.

As the ball is thrown upwards as it moves up, kinetic energy is converted to potential energy. The kinetic energy decreases and the potential energy increases. At the topmost point all the kinetic energy of the ball has been converted to potential energy.

When the ball starts to fall down again, the stored potential energy is converted to kinetic energy.

When the object is thrown up the kinetic energy is (1/2)*m*v^2, this is converted to potential energy which is equal to m*g*h, where h is the height of the highest point it reaches.

When it falls (1/2)*m*v^2 = m*g*h when it reaches the point it was thrown from. Hence the velocity is 7.5 m/s again, only the direction is opposite to that at which it was thrown upwards.

Here the ball falls down in the same vertical path. So it acquires the velocity of equal manitude but in the opposite direction while reaching the starting point compared to its projected velocity vertically upward.

The motion of the ball from the start is given by:

V = u+at. where u is the starting velocity and v is the final velocity, g is the scceleration due to gravity, s is the distance travelled.

The thown up ball goes up for a time t, till its upward velocity v becomes zero due to the downward pull of acceleration due to gravity.

So v = 0 = u+at = 7.5m/s-9.8t. Or t = 7.5/9.8 sec.

During this time the ball has reached height of Ss= ut -(1/2) at^2 = 7.5(7.5/9.8) - (1/2)(9.8)(7.5/9.8)^2 = (7.5)^2/19.6= 2.8699m nearly.

Nowe the ball traces back with the same path and same distance.

So we aply v^2-u^2= 2gs, where v = initial velocity at the height = 0 and v = final velocity whils reaching starting point.. ans s = (7.5)^2/(2g).

So v^2-u^2 = 2gs.

v^2 = 2gs , a u = 0.

v = 2g (7.5)^2/(2g)

v^2 = 7.5^2

v = 7.5 m/s

So the velocity of the ball while reaching back the starting point is 7.5m/s downward.

Assuming that the ball faces no air resistance, the ball's velocity, when it returns to the original point, will be equal and opposite to the starting point. So as the starting velocity was +7.5 m/s - that is, 7.5 m/s in upward direction - the velocity when it returns to the starting point will be 7.5 m?s in down ward direction, or -7.5 m/s.

The velocity of any object thrown vertically up gets reduced by the force of gravitation. This results in downward acceleration of about 9.8 m/s. The object continues to rise till its velocity becomes 0. At this point the object starts fall down, under the influence of gravitational force, with the same downward acceleration of 9.8 m/s. During this fall, the velocity of the abject at any given height is equal and opposite of the velocity at that point during upward rise.