A base ball is through with an initial velocity of 1/2m/s that makes on angle of 53 degree the horizontal (take g=10m/s square) a. The time required to reach maximum hight?
b. The time for with the stays in air?
c. The maximum hight reached by the ball?
d. the hight of the ball after 4second?
e. The vertical & Horizontal components of its velocity just after 4second?
f. The magnitude of its velocity b just after 4second?
g. The range of projection?

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Distance covered = speed x timethe distance traveled going against the stream = distance traveled with the stream.therefore it is the speed and the time that are the variables.as he travels the same distance both ways, speed x time is the same for each way.going upstream his speed is 4 kph; going downstream his speed is 6 kph.therefore we have two simultaneous equations: time upstream x 4 = time downstream x 6time upstream + time downstream = 60 minutestherefore we have the equation: time upstream = 6/10 hour, time downstream therefore = 4/10 hour.distance traveled at 4km/hr in 6/10 hour = 24/10 = 2.4km.check: distance traveled at 6km/hr in 4/10 hour = 24/10 = 2.4km.