A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.20 m/s at an angle of 17.0° below the horizontal. It strikes the ground 6.00 s later.
(a) How far horizontally from the base of the building does the ball strike the ground?

m

(b) Find the height from which the ball was thrown.

m

(c) How long does it take the ball to reach a point 10.0 m below the level of launching?

Question

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.20 m/s at an angle of 17.0° below the horizontal. It strikes the ground 6.00 s later.
(a) How far horizontally from the base of the building does the ball strike the ground?
  

 m

(b) Find the height from which the ball was thrown.
  

 m

(c) How long does it take the ball to reach a point 10.0 m below the level of launching?

jfraser · Answer

break the velocity vector into horizontal and vertical components

solve each one independently using the velocities and time

irishboy123 · Answer

a) horizontal velocity = 8.2cos(17) = 7.84m/s
horizontal distance travelled in 6 secs = 6 * 7.84 = 47m

b) vertical velocity = 8.2sin(17) = 2.4m/s
x = ut + 1/2a(t*t)
= 2.4*6 + 1/2(9.8)(6*6) = 190.8m

c) you could use x = ut + 1/2at^t BUT to avoid more painful quadratic solve it this way:
vertical velocity after it has dropped 10 m given by:
v^2 = u^2 + 2as
v^2 = 2.4*2.4 + 2(9.8)(10) = 201.8
v = 14.2m/s
then from v = u + at
t = (v-u)/a = (14.2-2.4)/9.8 = 1.2s.