The height (in meters) of a projectile shot vertically upward from a point 4 m above ground level with an initial velocity of 25.5 m/s is h = 4 + 25.5t − 4.9t2 after t seconds. When does the projectile meet its maximum height and what is its maximum height?

A projectile is launched straight up in the air. Its height​ (in feet) t seconds after launch is given by the function f(t)=−16t^2+407t+4. Find its average velocity from 2 seconds to 10.6 seconds.

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of vo = 23.0 m/s. The cliff is h = 61.0 m above a flat, horizontal beach as shown in the figure. h (a) What are the coordinates of the initial position of the stone? X0 = m Yo = (b) What are the components of the initial velocity? 0x m/s Voy* m/s (c) Write the equations for the x- and y-components of the velocity of the stone with time. (Use the following as necessary: t. Let the variable t be measured in seconds. Do not include units in your answer.) Vx= Vy= (d) Write the equations for the position of the stone with time, using the coordinates in the figure. (Use the following as necessary: t. Let the variable t be measured in seconds. Do not state units in your answer.) X= y = (e) How long after being released does the stone strike the beach below the cliff? S (f) With what speed and angle of impact does the stone land? m/s ° below the horizontal

A projectile is thrown from the top of a cliff with an initial speed of 25 m/s at an angle of 600 up from the horizontal. If the projectile lands 100 m from the base of the cliff how tall is the cliff?