# A ball of mass 0.50 kg is thrown straight up into the air at a speed of 24 m/s from a heightof 32 m. a) Find I) the maximum height of the ball ii) the speed at ground level iii) thespeed at 16 m...

A ball of mass 0.50 kg is thrown straight up into the air at a speed of 24 m/s from a height
of 32 m. a) Find I) the maximum height of the ball ii) the speed at ground level iii) the
speed at 16 m above the ground.

ncchemist | Certified Educator

You can only ask one set of questions at a time so I edited down accordingly.  At its highest position, the velocity of the ball will be 0.  The equation need for the first part is:

r = (vo)^2 / 2g

where r is the final position, vo is the inital upward velocity, and g is the acceleration due to gravity (9.8 m/s^2).  Plugging in the values:

r = (24 m/s)^2 / 2*(9.8 m/s^2) = 29.4 m

This means that the ball will travel upward for 29.4 meters before starting to fall back down.  Since the initial height of the ball was 32 m, then the maximum height of the ball is 32 m + 29.4 m = 61.4 m.

The velocity of the ball at ground level means that the ball will have fallen downward 61.4 m from an initial velocity of 0 m/s at the maximum height.  The equation needed for the second and third part is:

v^2 = vo^2 + 2gr

where v is the final velocity, vo is the initial velocity, g is the same acceleration due to gravity, and r is the distance travelled in free fall.  Since the velocity of the ball is 0 at the maximum height of 61.4 m, then the equation simplifies to:

v^2 = 2gr

When the ball reaches ground level it will have falled a distance of 61.4 m.

v^2 = 2(9.8 m/s^2)(61.4 m) = 1203.44 m^2/s^2

v = sqrt(1203.44) = 34.7 m/s

At a height of 16 m above the ground, the ball will have fallen a total of 61.4-16 = 45.4 m.

v^2 = 2(9.8 m/s^2)(45.4 m) = 889.84 m^2/s^2

v = sqrt(889.84) = 29.8 m/s

So the ball will reach a maximum height of 61.4 m, at ground level the speed will be 34.7 m/s, and at 16 m above the ground the speed will be 29.8 m/s.