A ball is shot straight up into the air from the ground with initial velocity of 44ft/sec44ft/sec. assuming that the air resista
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Answer:
a. 6.69m/s
b. y=4.48m
c. t=1.43secs
Explanation:
Data given, acceleration,a=35m/s^2
distance covered,d=64cm=0.64m,
a. to determine the speed, we use the equation of motion
initial velocity,u=0m/s
if we substitute values we arrive at
b. After taking the shot,the acceleration value is due to gravity i.e a=9.81m/s^2
and the distance becomes (y-2.2) above the ground. When it reaches the maximum height, the final velocity becomes zero and the initial velocity becomes 6.69m/s.
Hence we can write the equation above again
if we substitute values we have
c. the time it takes to arrive at 1.83m is obtain by using the equation below
if we insert the values, we solve for t , hence t=1.43secs