physics question

Question

From the top of a building, a physics student tosses a ball, which is given an initial velocity of 8.50 m/s at an angle of 15.0 below the horizontal. It strikes the ground 5.00 s later.

(a) How far horizontally from the base of the building (vertically below where the ball is thrown) does the ball strike the ground?

(b) Find the height from which the ball was thrown.

(c) How long does it take the ball to reach a point 10.0 m below the level of launching?

Answer 1

Ball is thrown at 8.5 m / sec at an angle of 15 degrees below horizontal. Ball reaches ground 5 seconds after it was thrown

Step 1 resolve velocity into horizontal and vertical components

Horizontal Component = 8.5 * cos(15) = 8.21 m / sec

Vertical Component = 8.5 * sin(15) = 2.2 m / sec

time = 5 seconds

acceleration due to gravity is 9.8 meters per second per second

a) horizontal distance travelled = horizontal velocity * time = 8.21 * 5 = 41.05 meters

b) vertical distance travelled = vertical velocity * time + 0.5 * acceleration * time * time

2,2 * 5 + 0.5 * 9.8 * 25 = 133.5 meters

c) s = vertical velocity * time + 0.5 * 9.8 * time * time

10 = 2.2 * time + 4.9 * time * time

4.9 * time * time + 2.2 * time - 10 = 0

time * time + 0.45 * time - 2.04 = 0

a = 1

b = 0.45

c = -2.04

Applying quadratic formula

time = (-0.45 +- sqrt(0.45*0.45 - 4 * -2.04) ) / 2

time = (-0.45 +- sqrt(8,36))/2

time = (-0,45 +- 2.89) / 2

time = (-0.45 + 2.89) / 2 = 2.44 / 2 = 1.22 seconds