Calculate maximum height

Question

A bullet is fired upwards initially with a velocity of 300m/s. Calculate the following: a) Maximum height reached by the bullet. b) Time taken to reach maximum height. c) Velocity of the bullet when it is 200m from the ground during downward movement.

Answer 1

???????????? up ?????????? du yu meen vertikal ????

????????????? "velosity"??????????   maebee yu meen SPEED=300 meter/sekond

????? yu somehow put gun on the grown & get it tu point strate up???????

me assume yu want GRAVITY tu sloe down the slug

vertikal speed=300 - gravity....32.2 ft/sek^2 or 9.780495 meter/sek^2

so speed=0 wen 9.780495x=300, x=300/9.780495=30.673294 sekonds

average speed=(start-speed -end-speed)/2=(300+0)/2=150

distans it go up=speed*time=150*30.673=4600.9994 meters

yu shood du sum av the werk, so me leev last quesshun tu yu

Answer 2

s=ut-gt^2/2 is the equation of motion where g=9.81 m/s/s is the approximate acceleration of gravity and u=initial speed=300m/s.

This is the equation of a parabola and the maximum height is the vertex. At this point the vertical speed is zero as the bullet slows down under gravity and then reverses direction. The rate of change of the height is the speed and this is u-gt. When this is zero u=gt so t=u/g=300/9.81=30.58 m/s. s=300*300/9.81-9.81*(300/9.81)^2/2=4587.2m approx. So the max height is about 4587m, and the time to reach that height is about 30.6 seconds.

When s=200, 200=300t-gt^2/2 so t^2-600t/g+400/g=0 from which t^2-600t/g+(300/g)^2=(300/g)^2-400/g; (t-300/g)^2=(300/g)^2-400/g. t-300/g=±√((300/g)^2-400/g)=±29.91 approx. Therefore t=300/g±29.91, t=60.49 seconds (downward path). The speed is 300-gt=-293.41m/s which means (because of the minus sign) that the bullet is travelling down at a speed of 293.41 m/s (approx).