KE before collision =1/2 m v2 = 1/2 m (15)2= 1/2 m 225 (Velocity)2 after collision v2=u2 +2as Final velocity is 0 when object is at maximum height u is the initial velocity just after the rebound a is due to gravity = -9.8ms-2 s is the displacement = 5m 0 = u2 + 2(-9.8)5 u2=98 (ms-1)2 KE after rebound = 1/2 m 98 since mass does not change Fraction of KE remaining = 98/225 Fraction lost 1 - fraction remaining 1- (98/225) = 127/225 Not: if g is taken to be 10ms-2, then fraction is 125/225 = 4/9
answer= 1.06kg*m/s
max height = zero
because its not moving, so m*v=0
then,
v^2=Vinitial^2+2ad
Vfinal^2=Vinitial^2+2ad
0=225-19.6*d
Vinitial=15
Vfinal= 0
a= -9.8
d=225/19.6=11.47m
halfway = 5.37m
soooo
V^2=15^2+2*-9.8*5.73 (to know half)
v^2=122.692
v= sqrt (112.692)
= 10.6m/s
p=mv
p=(0.1)(10.6)=1.06kg*m/s
By: lgmena
OR
V = (15) √0.5
= 15/1.414
= 10.6 m/s
P= mv
= 1.06 kg m/s
initial KE= final GPE
1/2 m v^2 = mgh
V= sqrt(2gh)
V=sqrt(2 x 9.80m/s^2 x 5m)
V=9.90 m/s upwards
A ball is thrown horizontally from a cliff at a speed of 15 m/s and strikes the ground 45 meters from the base of the cliff. How high was the cliff (rounded to the nearest meter)?
Answer: 44 meters
yes, i just test right now
If it was thrown horizontally or dropped, and hit the ground 3.03 seconds later, then it hit the ground moving at a speed of 29.694 meters (97.42-ft) per second. If it was tossed at any angle not horizontal, and hit the ground 3.03 seconds later, we need to know the direction it was launched, in order to calculate the speed with which it hit the ground.
False, provided the drop occurs no sooner than the throw, and the ground is flat .
No. They both hit the ground at the same time. This is because the VERTICAL component of velocity in both cases is the same.
No. They both hit the ground at the same time, because the VERTICAL component of velocity in both cases is the same.
Answer: 44 meters
Answer: 3 seconds
The ball was thrown horizontally at 10 meters per sec, and the thrower's arm was 78.4 meters above the base of the cliff.
10 m/s
64 METERSA+
64 metersIf a ball is thrown horizontally at 20 m/s from the top of a cliff that is 50 meters high, the ball will strike the ground 64 m from the base of the cliff (20m/s x 3.2 s).
"3.2" or "3.20" please put all of that
It doesn't matter whether the object is thrown down, up, horizontally, or diagonally. Once it leaves the thrower's hand, it is accelerated downward by an amount equal to acceleration of gravity on the planet where this is all happening. On Earth, if you throw an object horizontally, it accelerates downward at the rate of 9.8 meters per second2 ... just as it would if you simply dropped it. Whether it's dropped or thrown horizontally, it hits the ground at the same time.
yes, i just test right now
if the bal is thrown by making 45 degree angles. with the ground..it will travel maximum distance...
They should reach the ground together, since their initial vertical speed is the same, namely zero.
If it was thrown horizontally, it had an initial velocity of 10 meters/sec parallel to the ground. (It traveled 40 meters in 4 secs with no acceleration. x=vt) It also took 4 secs to travel vertically. It started with a vertical velocity of 0 m/s. Using x=v0 + (1/2) a t2 a = -g ( Acceleration due to gravity 9.8m/s2) x=0-(1/2)g*16 = -8 * 9.8 = -78.4 m It fell 78.4 meters before coming to a stop.