Question

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.55 m/s. If the roof is pitched at 22.0° below the horizon and the roof edge is 2.90 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground. HINT Solve for the time of flight using vy2 = v0y2 − 2gΔy and then vy = v0y − gt. Solve for the horizontal distance using the horizontal displacement equation. Click the hint button again to remove this hint. (a) the time the baseball spends in the air (in s) 0.434 Incorrect: Your answer is incorrect. s (b) the horizontal distance from the roof edge to the point where the baseball lands on the ground (in m) m

Answer 1

Answer:

Explanation:

The time that the baseball spends in the air is known as time of flight and it is expressed as shown;

T = Using(theta)/g where;

U is the velocity of the baseball

g is the acceleration due to gravity.

Given parameters

U = 4.55m/s

theta = 22.0°

g = 9.81m/s²

Substituting the values in the formula;

T = 4.55sin22°/9.81

T = 4.55(0.3746)/9.81

T = 1.7045/9.81

T = 0.1738second

Hence the time flight of the baseball is 0.1738second

b) The horizontal distance covered by the ball is called the RANGE in projectile.

Range = U√2H/g

U = 4.55m/s

H is the maximum height = 2.90m

g = 9.81m/s²

Substitute the given parameters into the formula

Range = 4.55√2(2.90)/9.81

Range = 4.55√5.8/9.81

Range = 4.55√0.5912

Range = 4.55(0.7689)

Range = 3.4986m

Hence the horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.4986m