Question

5. If a ball is thrown directly upward with a velocity of 46 ft/s, its height (in feet) aftert seconds is given by y = 46t - 16t? What is the maximum height attained by the ball? Round your answer to the nearest whole number.

Answer 1

Given: u=46ft/sec (initial velocity)

y(t)=46t-16t² (height in feet after t seconds)

Now, the final velocity, say v(t) is given by-

v(t) = f(y(t)) = 1 (46t – 16t?) = 46 – 32

Also,  

alt) = 1 = (46 - 32t) = -32

-ve sign of acceleration a(t) indicate that ball is thrown upward(against the direction).

Now, since acceleration a(t) is a constant and -ve implies that velocity will decrease with time.

Therefore, maximum height is attained when final velocity reaches to 0.

Hence, from v(t) equation we get -

46 – 32t = 0 +t= 46 sec

Putting the value of 't' in y(t) we get the desired maximum height -

90 ) = 46/ ) - 16() = 33.0625ft

Since, we need to find the answer in nearest whole number, therefore the answer will be 33 ft.