A horizontal projectile is shot with an initial velocity of 25.2 m/s from a height of 8.43 m above the ground. Neglecting air-resistance, what is the magnitude of the final velocity (m/s)?

You threw a tennis ball from a balcony 4.1 m above the ground upward at 45 degrees above horizontal with an initial speed of 4.0 m/s. What was the magnitude of the total final velocity?

What is the range in meters of a projectile launched at an angle of θ1 = 78.3° with an initial speed of 58 m/s?

A horizontal projectile is shot with an initial velocity of 9.9 m/s from a height of 5.8 m above the ground. Neglecting air-resistance, what is the horizontal distance of the landing point (measured in meters from the base of the shooting point)?

You throw a ball from the ground level at an angle of 0 = 40.0° above the horizontal. The ball lands on top of a cliff that has a height of h=10.0 m. When you release the ball, it is traveling at a speed of 25.0 m/s. How long does it take for the ball to land on top of the cliff?

A projectile is launched with a launch angle of 30° with respect to the horizontal direction and with an initial speed of 26 m/s.    How do the vertical and horizontal components of the projectile's velocity vary with time?The initial velocity in the x-direction vx0 is related to the initial speed by vx0 = v0 cos 30°. The constant velocity in the x-direction means that the equation describing the time dependence of x for the particle, with x0 taken as 0, is x = x0 + vx0t = 0 +       m/s    t. The equation for the vertical coordinate, which is constantly accelerating downward at g = 9.8 m/s2, is y = y0 + vy0t −  1 2 gt2 =      m/s    t +      m/s2    t2.

Consider a projectile launched at a height h feet above the ground and at an angle ? with the horizontal. If the initial velocity is v0 feet per second, the path of the projectile is modeled by the parametric equations x = t(v0 cos(?)) and y = h + (v0 sin ?)t - 16t2.Let h=6, v0=63.25, and ?=0.79.  What is the range of the projectile (i.e., how far does it travel? Round your answers to two decimal places).

Imagine an archer preparing to fire an arrow. The archer aims the arrow directly ahead, that is to say, parallel with the ground. Bows are drawn to the corner of the mouth when fired. Let's say, for this particular archer, that puts the arrow 1.8 m (about 5.5 feet) above the ground. Assume that we have measured the initial speed of arrows fired by this archer with this bow and it is 100 m/s. (Ignore wind resistance) a) At what point during the arrow's flight will it be moving the fastest? (Be careful here, think about it. You don't need to do any calculations to figure this out.) b) How long will it take for the arrow to hit the ground?c) How far will the arrow travel during this time period?

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of vo= 13.0 m/s. The cliff is h = 47.0 m above a flat, horizontal beach as shown in the figure. (a) What are the coordinates of the initial position of the stone? Xo = Yo= (b) What are the components of the initial velocity? Vox m/s m/s Voy Vx= (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.) = m m X= (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.) y =

A projectile is launched from and returns to ground level, as the figure shows. There is no air resistance. The horizontal range of the projectile is measured to be R= 177 m, and the horizontal component of the launch velocity is vox = 22.0 m/s. Find the vertical component voy of the projectile. +v voy Var=+22.0m/s to +x Number i R=177 m Units