A ball is thrown vertically upward at 10m/s. How high will it get, how long will it be in the air, and how fast will it be moving when it hits the ground? Step 1: Write the given Initial velocity (upward)= 10 m/s Step 2: Determine the formula/s that may be utilized For upward direction: For downward direction: Step 3: Find the maximum height of the ball Use the eq 3 for upward direction In order to find hte max height, use 0 m/s as the final velocity (0 m/s)²=(10 m/s)²-2(9.81 m/s²)(max height) 10²=2*9.81*max height The maximum height of the ball will reach is 5.10 meters. Step 4: Find the time it will stay in the air For upward motion, use equation 1 0 m/s= 10 m/s-(9.81 m/s²)(t upward) The time it will take to reach the maximum height is 1.02 sec. For downward motion, use equation 2 5.10 meters=(0 m/s)(t downward)+(1/2)(9.81 m/s²)(t downward)² The time it will take to reach the ground from maximum height is also 1.02 sec. The total time the ball will be in the air ...
Write an equation in which the quadratic expression 2x^2-6x-8 equals 0. Show the expression in factored form, find the solution and explain what your solutions mean for the equation. Remember to show your work. Answer:(x-4)(x+1) Step-by-step explanation: 2x^2-6x-8 is given as 0 So 2x^2-6x-8=0 2(x^2-3x-4)=0 That is x^2-3x-4=0 It can also be written as x^2+(-4x+x)-4=0 Taking x common x(x-4)+1(x-4)=0 Taking x-4 common (x+1)(x-4)=0 That is Either x+1=0 or x-4=0 Therefore x=-1 or x=4 The significance is that if we put these values in the equation then the solution will be 0
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