Question

1) A ball is thrown from the top of a roof at an angle of 20 o with respect to the vertical. 1 s later a ball is dropped from the top of the roof

a) If the height of the roof is 20 m, determine the velocity with which the first ball must be thrown in

order for both balls to land at exactly the same time.

b) Imagine the velocity of the first ball is now known, and instead it is the height of the building which

is unknown. Can you obtain an equation for the height of the building which would result in both balls

landing on the floor at the same time

Answer 1

a) consider the motion of ball dropped ::

distance travelled = d = - 20 m

initial velocity = V_i = 0 m/s

acceleration = a = g = -9.8 m/s²

time taken = t = ?

Using the kinematics equation ::

d = V_i t + (0.5) a t²

inserting the values

- 20 = 0 x t + (0.5)(-9.8) t²

t = 2.02 sec

motion of the ball thrown ::

initial velocity in up direction = V_i = V cos20

displacement = d = -20 m

acceleration = a = g = -9.8 m/s²

time taken = t = 2.02

Using the kinematics equation ::

d = V_i t + (0.5) a t²

inserting the values

- 20 = V cos20 x (2.02) + (0.5)(-9.8) (2.02)²

V = -0.00318 m/s

b)

consider the motion of ball dropped ::

distance travelled = -d

initial velocity = V_i = 0 m/s

acceleration = a = g = -9.8 m/s²

time taken = t

Using the kinematics equation ::

-d = V_i t + (0.5) a t²

inserting the values

-d = 0 x t + (0.5)(-9.8) t²

t = sqrt(d/4.9)

motion of the ball thrown ::

initial velocity in up direction = V_i = V cos20

displacement = d

acceleration = a = g = -9.8 m/s²

time taken = t

Using the kinematics equation ::

d = V_i t + (0.5) a t²

inserting the values

- d = V cos20 x (sqrt(d/4.9)) + (0.5)(-9.8) (sqrt(d/4.9))²

by inserting the value of ''v'' we can calculate ''d''