A ball is ejected on a top of a 25 m hill, with a velocity of 30 m/s, in a direc

Question

A ball is ejected on a top of a 25 m hill, with a velocity of 30 m/s, in a direction of 30 degrees above horizontal. Ignore air resistance. Use g = 10 m/s^2. (a) How long does it take for the ball to reach the maximum height? (b) What is the maximum height the ball can reach relative to flat area? (c) How long does it take for the ball to reach the ground after passing the maximum height? (d) What is the maximum horizontal distance it can travel? (e) What is the ball's impact velocity (Magnitude only)?

Explanation / Answer

a)

consider the motion in vertical direction

Voy = initial velocity = 30 Sin30 = 15 m/s

Vfy = final velocity at highest point = 0 m/s

a = acceleration = - 10 m/s2

t = time taken

Using the equation

Vfy = Voy + at

0 = 15 + (-10) t

t = 1.5 sec

b)

h = total height from ground

using the equation

V2fy = V2oy + 2a(h - 25)

02 = 152 + 2 (-10) (h - 25)

h = 36.25

c)

T = total time to reach the ground

using the equation

d = Voy T + (0.5) a T2

- 25 = 15 T + (0.5) (-10) T2

T = 4.2 sec

time taken after maximum height = 4.2 - 1.5 = 2.7 sec

d)

Horizontal distance travelled = X

Vox = velocity = 30 Cos30

X = Vox T

X = (30 Cos30 ) (4.2) = 109.12 m

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