1.) Auston’s racecar has a speed of 80 m/s when Auston releases a drag parachute. If the parachute causes a deceleration of -4 m/s2, how far will the car travel before it stops?
2.) Jordan throws a golfball straight up from the top of a building 50 meters high. The initial speed of the golfball is 20 m/s and it just misses the edge of the roof when it eventually falls. Neglecting air resistance, find the velocity of the golfball just at the instant it hits the ground.
a. vf^2=vi^2+2ad solve for d
b. KEfinal=KEinitial+PE initial
1/2 m vf^2=1/2 m 20^2 + mg*50
solve for vf
1.) To find the distance the car will travel before it stops, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (which is 0 m/s when the car stops)
u = initial velocity (80 m/s for the car)
a = acceleration (deceleration in this case, -4 m/s^2)
s = distance traveled
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Substituting the values, we get:
s = (0^2 - 80^2) / (2 * -4)
s = (-6400) / (-8)
s = 800 meters
Therefore, the car will travel 800 meters before it stops.
2.) To find the velocity of the golf ball just at the instant it hits the ground, we need to consider the motion of the ball when it is thrown up and when it falls down.
When the ball is thrown up, its initial velocity is 20 m/s, and the acceleration due to gravity is -9.8 m/s^2 (negative because it is acting in the opposite direction of the ball's motion). We can use the following equation to find the time it takes for the ball to reach its maximum height:
v = u + at
where:
v = final velocity (0 m/s when the ball reaches maximum height)
u = initial velocity (20 m/s for the ball)
a = acceleration (gravity, -9.8 m/s^2)
t = time taken
Rearranging the equation, we have:
t = (v - u) / a
t = (0 - 20) / (-9.8)
t = 2.04 seconds (rounded to 2 decimal places)
So, it takes approximately 2.04 seconds for the ball to reach its maximum height.
To find the velocity of the ball just at the instant it hits the ground, we can use the equation:
v = u + at
where:
v = final velocity (which we are trying to find)
u = initial velocity (0 m/s when the ball starts falling down)
a = acceleration (gravity, -9.8 m/s^2)
t = time taken (which is twice the time it took for the ball to reach its maximum height, because the motion is symmetrical)
Substituting the values, we get:
v = 0 + (-9.8) * 2.04
v = -19.92 m/s
Therefore, the velocity of the golf ball just at the instant it hits the ground is approximately -19.92 m/s (negative because it is in the downward direction).