A ball is projected with a horizontal velocity of #5.1# #ms^-1# at a height of #2.5 xx 10^2# #m#. How far does it travel before it hits the ground?

1 Answer
David G.
Jun 24, 2017

The horizontal distance traveled will be #s=u_xt=5.1xx7.1=36.2# #m#.

Explanation:

As for all projectile motion questions, we can treat the horizontal and vertical direction independently. If we ignore air resistance, the horizontal velocity will remain constant at #u_x=1.5# #ms^-1#.

We need to calculate the time taken for the ball to fall. The initial velocity #u_y# in the vertical direction is #0# #ms^-1#.

#s = u_yt +1/2a_yt^2#

#t = sqrt((2s)/a)# (since #u_y=0# and we can rearrange to make #t# the subject)

#t = sqrt((2(2.5xx10^2))/9.8)=7.1# #s#

The horizontal distance traveled in this time will be #s=u_xt=5.1xx7.1=36.2# #m#.

Related questions
See all questions in Displacement and Velocity
Impact of this question
1486 views around the world
You can reuse this answer
Creative Commons License