Help! Please?

A projectile is thrown from the top of a building with initial horizontal velocity #4 m/s#. The building is 45m high. What is the velocity of the projectile after 2 seconds.

I make it #20 m/s#, but the options are:

#4 m/s#

#8 m/s#

#40 m/s#

#80 m/s#

I keep recalculating according to how you are supposed to solve these, but still get the same answer. I don't no what else to do.

Thanks.

1 Answer
May 4, 2018

Please see the explanation below.

Explanation:

Here is what I think about this problem :

The horizontal velocity will remain constant.

#u_x=4ms^-1#

Let the acceleration due to gravity #g=9.8ms^-2#

The vertical component of the velocity is #=v_y#

After #t=2s#

#v_y(2s)=0+ g t =9.8*2= 19.6ms^-1#

The velocity is

#v(t)=sqrt(v_x^2+ v_y^2)=sqrt(19.6^2+4^2)=20ms^-1#

I get the same answer as you. There is a mistake either in the question or in the options.

Good Luck!