The stone has already hit the ground. The problem assumes (and it should have st
ID: 585512 • Letter: T
Question
The stone has already hit the ground. The problem assumes (and it should have stated) that if the stone hits the ground, it will stay at that spot from that time onward. So you have to check to see if the given time is greater than the total time the stone will be in the air; if it is, then the stone has already hit the ground., and if not then the stone is still in the air.
A stone is catapulted at time t = 0, with an initial velocity of magnitude 20.2 m/s and at an angle of 38.8° above the horizontal. what are the magnitudes of the (a) horizontal and (b) vertical components of its 5.19 s. Assume that the displacement from the catapult site at t 1.18 s? Repeat for the (c) honzontal and d) vertical components at t = 1.72 s, and for the (e) honzontal and f) vertical components at t catapult is positioned on a plain horizontal ground Units Tm Units Tm Units Tm (a) Number 5703 (b) Number 8.108 (c) Number 27.077 (d) Number 7.2647 (e) Number 81.7 (f) Number -66.385 Uns Unis Units Unit UnitExplanation / Answer
(a) The horizontal component of the velocity is
20.2 m/s cos(38.8 deg) = 15.7426 m/s
After 1.18 s, the horizontal displacement is 18.576 m
(c) - I'll get back to (b) --
After 1.72 s, the horizontal displacement is 27.078 m
(b) The vertical component of the initial velocity is
20.2 m/s sin(38.8 deg) = 12.657 m/s,
and this decreases by 9.8 meters per second per second.
y = 12.657 t - (1/2) 9.8 t^2
When t = 1.18 s, y = 8.1125 m
(d) Using the formula developed in (b),
when t = 5.19 s, y = -66.29 m
Please note the negative sign depicts that the stone has already hit the ground, therefore the stone cannot trace the path anymore and will continue to stay at that spot from that time onwards.
Calculstions maybe faulty, logic is correct.
Hope this helps :)
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