Question

Suppose you throw a ball at 20 m/s at an angle of 60 degrees from the horizontal from the top of a 40 m building.

a. what is the maximum height?

b. what is the velocity at the highest point?

c. what is the final velocity just as it hits the ground?

d. how far did it travel in the x direction when it hit the ground?

Answer #1

Here ,

u = 20 m/s

theta = 60 degree

h = 40 m

a) maximum height = 40^2 + (20 * sin(60 degree))^2/(2 * 9.8)

maximum height = 1615 m

b) velocity at the highest point = 20 * cos(60 degree)

velocity at the highest point = 10 m/s

c) final velocity at the ground

v^2 - 20^2 = 2 * 9.8 * 40

v = 34.4 m/s

the final velocity at ground is 34.4 m

d) let the distance is x

0 = 140 + x * tan(60 degree) - 9.8 * x^2/(2 * (20 * cos(60 degree))^2)

solving for x

x = 74 m

the horizontal distance traveled is 74 m

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