First let's find the time it takes for the first ball to land:
Acceleration is a=-g so vertical velocity is V=-gt + V1sin(30).
Position is thus
S=(-1/2)gt^2 +V1t sin(30).
Solving for t gives
t=2V1sin(30)/g
The second ball has the same position function except for the new velocity, which is given by
V2=2V1. Putting this in and solving for t2 gives
t2=4V1sin(30)/g.
It takes twice as long for the second ball to land on the ground.
The horizontal distance of ball 1 is S1 = V1t cos(30). Again we look at ball 2's distance by substituting V2=2V1 and get
S2 = 2V1t2 cos(30).
Note here I put in t2 since it will fly for that amount of time. But we already saw that
t2 = 2t1
So S2=4V1 cos(30)
That is the second ball goes 4 times further than the first one. This is because it is going twice as fast along both the horizontal and the vertical. It moves horizontally twice as fast for twice as long.
the missing force is spring force.
The object is hanging from the spring and the spring is stretched by some distance from its equilibrium position. due to this stretch in the spring , a spring force starts acting on the object trying to regain its equilibrium position.
the spring force is given as
F = kx
where F = spring force ,k = spring constant , x = stretch in the spring.
the spring force balances the weight of the object in down direction and hence keeps the block from falling down.
At 100 km/hr, the car's kinetic energy is
KE = (1/2) (mass) (speed)²
KE = (1/2) (1575 kg) ( [100 km/hr] x [1000 m/km] x [1 hr/3600 sec] )²
KE = (787.5 kg) (27.78 m/s)²
KE = 607,639 Joules
In order to deliver this energy in 2.9 seconds, the engine must supply
(607,639 J / 2.9 sec) = 209,531 watts
<em>Power = 281 HP</em>
Ptolemy life is probably my most recommenced
Look at the third one i think its the answer
