A Chevy Camaro drives straight off the top level off a parking garage at 13 m/s. If the car landed 90 meters away from the base of the parking garage, how high (height) was the top level?

H = 234.85m

Explanation:

Using the formula for calculating range to find the maximum height

Range R = U√2H/g

U is the speed = 13m/s

H is the max height

R is the distance covered in horizontal direction

g is the acceleration

Substitute the given values as shown;

90 = 13√2H/9.8

90/13 = √2H/9.8

6.923 = √2H/9.8

6.923² = 2H/9.8

2H = 6.923²  *  9.8

2H = 469.70

H = 469.70/2

H = 234.85m

Hence the top level was 234.85m high

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//