Question

Krissy throws a ball from a height of 2.00 m above the ground. She throws the ball with a speed of 3.00 m/s directed 50.0 degrees above the horizontal. Find: a.) the speed of the ball at landing b.) the time it takes the ball to land c.) the angle with which the ball impacts the ground d.) the horizontal range of the ball e.) the maximum height of the ball f.) the time it takes the ball to reach maximum height g.) the speed of the ball at maximum height *This physics uses trig and algebra to slove problems.

Answer 1

hi= 2 m

u = 3 m/s

theta = 50 degree

a) Using third equation of motion

v^2 - u^2 = 2 * g * hi

v^2 - 3^2 = 2 * 9.8 * 2

v = 6.94 m/s

the speed of the ball at landing is 6.94 m/s

b)

let the time taken is t

y = u * sin(theta) * t - 0.50 gt^2

-2 = 3 * sin(50 degree) * t - 0.50 * 9.8 * t^2

solving for t

t = 0.915 s

the time taken is 0.915 s

c) let the angle is theta

v * cos(theta) - u * cos(theta) = 0

6.94 * cos(theta) - 3 * cos(50) = 0

theta = 73.8 degree below horizontal

d) horizontal range = 3 * cos(50) * 0.915

horizontal range = 1.76 m

e) maximum height = 2 + (u * sin(theta))^2/(2g)

maximum height = 2 + (3 * sin(50 degree))^2/(2 * 9.8)

maximum height = 2.27 m

f)

time taken to reach the maximum height = u * sin(theta)/g

time taken to reach the maximum height = 3 * sin(50 degree)/9.8

time taken to reach the maximum height = 0.23 s

g) speed at maximum height = horizontal speed

speed at maximum height = u * cos(50)

speed at maximum height = 3 * cos(50)

speed at maximum height = 1.93 m/s