A ball is thrown upwards and reaches the ground once again after 4 sec: (a)calculate the height that the ball reaches (b)the total distance covered by the ball (c)the velocity with which the ball hits the ground

h(t)=h(0)+vi*t-1/2 g t^2

0=0+vi*4-4.9(4^2)
so you need the initial velocity, vi, or vi=4.9*4 m/s.
height: vi*2
distance: 2*vi*2
vf=-vi

Height=364.56m

To calculate the answers to the given questions, we can use the equations of motion for the ball's vertical motion.

(a) To calculate the height that the ball reaches:

The ball follows a parabolic path when thrown upwards. At the highest point, its velocity becomes zero. Using the equation for the vertical displacement, we have:

h = ut - (1/2)gt^2,

where:
h is the height reached by the ball,
u is the initial velocity,
g is the acceleration due to gravity (9.8 m/s^2), and
t is the time taken.

In this case, the time taken for the ball to reach the ground is 4 seconds. At the highest point, the velocity is zero. Therefore, the time taken to reach the highest point is half of the total time, which is 2 seconds.

Substituting the values into the equation, we get:

h = (0) * (2) - (1/2) * (9.8) * (2)^2
h = -19.6 meters.

The height reached by the ball is -19.6 meters (negative because the height is measured from the ground level).

(b) To calculate the total distance covered by the ball:

The total distance covered by the ball is the sum of the upward and downward distances. Since the ball reaches the ground again, the total distance covered is twice the height reached. Using the value calculated above, we have:

Total distance = 2 * |h|
Total distance = 2 * |-19.6|
Total distance = 39.2 meters.

The total distance covered by the ball is 39.2 meters.

(c) To calculate the velocity with which the ball hits the ground:

At the highest point, the velocity becomes zero. From there, the ball falls freely under the influence of gravity and reaches the ground with a velocity. Using the equation for vertical velocity, we have:

v = u - gt,

where:
v is the final velocity (velocity at impact),
u is the initial velocity,
g is the acceleration due to gravity, and
t is the time taken (4 seconds).

Initially, the ball is thrown upward with an initial velocity u. Therefore, we need to calculate the initial velocity at the highest point.

Using the equation for vertical displacement and rearranging, we have:

h = ut - (1/2)gt^2,
0 = u(2) - (1/2)(9.8)(2)^2,
0 = 2u - 19.6,
2u = 19.6,
u = 9.8 m/s.

So, the initial velocity (at the highest point) is 9.8 m/s.

Now, let's calculate the final velocity at impact:

v = 9.8 - (9.8)(4),
v = 9.8 - 39.2,
v = -29.4 m/s.

The velocity with which the ball hits the ground is -29.4 m/s (negative because it is directed downward).