A baseball hit with a vertical speed of 18m/s upward. What is the speed 2s later?

2 Answers
Feb 3, 2018

#-1.6 m/s#

Explanation:

#v = v_0 - g t#

#"(-"g"t because we take the + velocity upwards)"#

#"So here we have"#
#v = 18 - 9.8 * 2#
#=> v = -1.6 m/s#
#"The minus sign indicates that the velocity is downwards, so"#
#"the ball is falling after it reached the highest point."#

#g = 9.8 m/s^2 = " gravity constant"#
#v_0 = " initial velocity in m/s"#
#v = " velocity in m/s"#
#t = " time in seconds"#

Feb 3, 2018

#2 m/s#

Explanation:

Here,the ball goes up due to a given initial velocity,but the gravitational force opposes its motion and when the upward velocity becomes zero,it comes down due to gravity.

So,here we can use the equation, #v=u-g t# (where, #v# is the velocity after time #t# with an initial upward velocity #u#)

Now,putting #v=0# , we get #t=1.8# ,which means the baseball reaches its highest point in #1.8 s# and then starts falling down.

So,in #(2-1.8)s# it will have a velocity of #0.2*10 m/s# or #2 m/s# downwards.(using #v'=u'+g t# while falling,#u'=0# and here required time is #0.2 s#)

ALTERNATIVELY

Simply,put the given values in the equation, #v=u-g t#

So,you get, #v= -2 m/s# that means velocity will be #2 m/s# downwards,as we took upward direction to be positive in this equation.

So,speed is #2m/s# (omit the negative sign,as speed can't be negative)