A pilot drops a bomb from a plane flying horizontally. Where will the plane be located when the bomb hits the ground

The plane will be located directly above the bomb because they both have the same horizontal speed.

The height risen by the ball is 39.8 m.

Explanation:

Given;

total time spent in air by the ball, t = 5.7 s

time to rise to maximum height, t = 5.7 /2 = 2.85 s

The initial velocity of the ball is given by;

v = u - gt

where;

v is the final velocity at maximum height = 0

u is the initial velocity of the ball

0 = u - gt

u = gt

u = (9.8 x 2.85)

u = 27.93 m/s

The vertical height traveled by the ball is given by;

v² = u² - 2gh

where;

v is the final velocity of the ball at maximum height = 0

g is acceleration due to gravity = 9.8 m/s²

0 = u² - 2gh

2gh = u²

h = u² / 2g

h = (27.93)² / ( 2 x 9.8)

h = 39.8 m

Therefore, the height risen by the ball is 39.8 m.