how long will it take a shelf fired from a cliff at an initial velocity of 800 metres per second at an angle of 30 degree below the horizontal to reach the ground 150 metre belowJason control rules of baseball with the horizontal component of velocity of 25 metres per second

0.373 seconds. First, calculate the initial vertical velocity of the shell. 800sin(30) = 800*0.5 = 400 m/s Now the formula for the distance traveled is d = 400 m/s * T + 0.5A T^2 Substituting known values gives. 150 = 400 m/s * T + 0.5*9.80m/s^2 T^2 150 = 400 m/s * T + 4.9 m/s^2 T^2 Arrange as a quadratic formula 0 = 400 m/s * T + 4.9 m/s^2 T^2 - 150 4.9 m/s^2 T^2 + 400 m/s * T - 150 = 0 Now solve for T using the quadratic formula with a=4.9, b=400, and c=-150 The calculated value is 0.373 seconds. Is this value reasonable? Let's check. The initial downward velocity is 400 m/s. So 150/400 = 0.375 seconds. Since the actual time will be a bit less due to acceleration by gravity and since the total time is so short, there won't be much acceleration due to gravity, the value of 0.373 is quite reasonable.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

honestly bro aint no one got time to answer these dumb question just do it p***y

step-by-step explanation: