Part a can be solve using the equation of trajectory:
Y = x tana + (g*x^2)/ [2(V0^2)*(cos a)^2]
Where y is the height
X is the length
G is the acceleration due to gravity
Vo Is the initail velocity
a is the angle of trajectory
1.2 = 1.35 tan(0) +
(9.81*1.35^2)/ [2(V0^2)*(cos 0)^2]
Solve for V0 = 2.729 m/s
b. can be solve using the formula
v = sqrt(2gy)
= sqrt ( 2*1.2*9.81)
= 4.852 m/s going
down ( 0 degree from the horizontal)
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Answer:
A concave mirror has a focal length of. 10.0 cm. What is its radius of curvature? ... 20.0 cm. 62. An object located 18 cm from a convex mirror produces a virtual image 9 ... cm. 75 cm. 66. Find the image position and height for the object shown in ... 1 block 1.0 cm. Vertical scale: 2 blocks 1.0 cm. F. I1 hi. 1.0 cm di. 2.7 cm. O1.
Explanation:
Hope This Helps
I’m pretty sure it is dropping it horizontally at a slower speed. Sorry if I’m wrong.
Answer:
magnitude = 304.14 km/h
direction: 
West of North
Explanation:
The final plane's vector velocity will be the result of the vector addition of one pointing North of length 300 km/h, another one pointing West of length 50 km/h.
To find the magnitude of the final velocity vector (speed) we need to apply the Pythagorean theorem in a right angle triangle with sides: 300 and 50, and find its hypotenuse:
km/h
The actual direction of the plane is calculated using trigonometry, in particular with the arctan function, since the tangent of the angle can be written as:
So the resultant velocity vector of the plane has magnitude = 304.14 km/h,
and it points West of the North direction.
