An object is projected vertically upwards at a speed of 22ms-1 (g=9.8-2) calculate the maximum height reached by the object (the speed at the top is 0) and the time taken between the object being projected and it landing again.
The answer given in the book is 25 m and 4.5s but by my calculations I cannot achieve this - any help appreciated I'm probably missing something simple out!
2007-03-08 02:04:32 · 2 answers · asked by louise faz 1 in Science & Mathematics ➔ Physics
v² = v0² -2gx so
0 = 22² - 2*(9.8)*x and
x = 22²/(2*9.8) = 484/19.6=24.69 meters high
Since the time going up and coming down are identical, calculate the time coming down as
s = at²/2 so
t = √(2s/a) = √(2*24.69/9.8) = 2.244 seconds and twice that is 4.488 seconds in the air
HTH ☺
Doug
2007-03-08 02:13:36 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
the formula to calculate the maximum ht =initial velocity to the power of 2 divided by twice the gravity
the answer would be 24.6938 that is approimately 25 m
time=initial velocity divided by gravity*2
the time would be 4.489 s that is approx . 4.5 s
2007-03-08 10:13:01 · answer #2 · answered by satwik 2 · 0⤊ 0⤋