A projectile is fired from the ground, reaches a maximum height of 42.4 m and lands a distance of 57.4 m away from the launch point. What was the projectile s launch velocity?

A cannonball is shot with an initial speed of 30.0 m/s and 25.0º angle above horizontal from a height of 12.0 m. How far can the cannonball go horizontally when it lands on the ground, in meters? Use g = 10.0 m/s2. (Hint: Clearly label quantities along x- and y- components and write down all steps.)

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the court is the tennis ball when it leaves the racket? Ans: 1.83 m SUL des Bud Question 3. A marble is thrown horizontally with a speed of 19.2 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 39° with the horizontal. From what height above the ground was the marble thrown? Ans: 12.33 m lage vob Y Question 4. A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is 640 the hullot strikes the target.

(a) From a balcony a child fires a toy arrow down to the ground below. The arrow has initial speed 12 m/s and is aimed at an angle of 30° below the horizontal. The arrow takes 1.2 seconds to hit the ground. (i) Calculate the initial horizontal (✰) and vertical (ŷ) velocity components of the arrow. (ii) Calculate the height of the balcony. (b) A projectile is launched with an initial speed of 75 m/s at an angle of 52° above the horizontal. If it lands at the same height from which it was launched, calculate how long the projectile remains in the air.

A baseball player makes perfect contact with the ball, striking it 45° above the horizontal at a point1.25 m above the ground. The home run hit just clears the 3.00 m wall 125.0 m from home plate.Draw a diagram. With what velocity did the baseball player strike the ball?

The initial speed of a tennis ball is 57.5 m/s and the launch angle is ?i = 16°. Neglect air resistance.Question 1) What is the maximum height, h, of the tennis ball? in mQuestione 2) What is the range, R, of the tennis ball? in m info given: what angle results in the greatest height=90 For this 90° angle, the horizontal component of velocity is zero, so the initial velocity is completely in the vertical direction, resulting in greatest height Kato tries substituting ty,max for t, 0 for yi, and h for yf, and gets h =  vi2 sin2 ?i 2g   When ?i = 90°, sin2 ?i is maximum, so h is maximum what angle results in the maximum horizontal range=45°  I substituted 2vi sin ?i g into the expression for the horizontal component of velocity, (vi cos ?i)t, and got  R =  2vi2 sin ?i cos ?i g when ?i = 45°, 2?i = 90°, and sin 90° = 1, which is its maximum value.

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

A basketball is shot from an initial height of 2.40 m (Figure 1) with an initial speed v = 9 m/s directed at an angle θ = 35˚ above the horizontal. How far from the basket was the player if he made a basket? At what angle to the horizontal did the ball enter the basket?

An object on the ground is launched at 35ˍm/s at an angle of 61° above the horizontal. If the object lands on top of a 5.9ˍm tall hill, then what is the horizontal displacement of the object?