A baseball is thrown straight up at 15 m/s. How high will it go?

2 Answers
Gió
Apr 16, 2016

I found #11.5m#

Explanation:

We can use here the general relationship from kinematics:

#color(red)(v_f^2=v_i^2+2a(y_f-y_i))#

where:
#v_i# is the initial velocity#=15m/s#;
#v_f# is the final felocity which is zero in our case;
#a# is the acceleraton of gravity #g=-9.8m/s^2# (downward);
#y_f# is the height reached from the ground where #y_i=0#.

So we get:

#0^2=15^2-2*9.8*(y_f-0)#

and:

#y_f=(225)/(19.6)=11.5m#

ali ergin
Apr 16, 2016

#h_m=11,47 "m"#

Explanation:

enter image source here
#h_m="maximum height"#
#g=9,81 N/(kg)#
#v_i=15 m/s" initial velocity"#
#alpha=pi/2#
#sin (pi/2)=1#
#h_m=v_i^2/(2*g)#

#h_m=15^2/(2*9,81)#

#h_m=225/(19,62)#

#h_m=11,47 "m"#

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