Question

You throw a 2 kg ball from the top of a 30 m cliff with an initial velocity of 25 m/s at an angle of 50 degrees.

a. using energy considerations, what is the maximum height of the ball?

b. what is the magnitude of the velocity when it hits the ground?

c. now, using kinematics, what is the magnitude of the velocity when it hits the ground?

Answer 1

here,

mass , m = 2 kg

height , h = 30 m

initial velocity , u = 25 m/s

theta = 50 degree

a)

let the maximum height be Hmax

using conservation of energy

0.5 * m * (u * sin(theta))^2 + m * g * h = m * g * hmax

0.5 * ( 25 * sin(50))^2 + 9.81 * 30 = 9.81 * hmax

solving for hmax

hmax = 48.7 m

the maximum height is 48.7 m

b)

let the magnitude of final velocity be v

using conservation of energy

0.5 * m * u^2 + m * g * h = 0.5 * m * v^2

0.5 * 25^2 + 9.81 * 30 = 0.5 * v^2

solving for v

v = 34.8 m/s

the magnitude of final velocity is 34.8 m/s

c)

let the final vertical speed be vy

vy^2 - uy^2 = 2 * h * g

vy^2 - (25 * sin(50))^2 = 2 * 30 * 9.81

solving for vy

vy = 30.9 m/s

vx = u * cos(theta) = 16.1 m/s

the magnitude of final velocity , |v| = sqrt(vx^2 + vy^2)

|v| = 34.8 m/s