A projectile is fired with an initial speed of 36.8 m/sm/s at an angle of 42.8 ∘∘ above the horizontal on a long flat firing range. Determine the maximum height reached by the projectile.   Determine the total time in the air.   Determine the total horizontal distance covered (that is, the range).

A quarterback is set up to throw the football to a receiver who is running with a constant velocity vecter vr directly away from the quarterback and is now a distance D away from the quarterback. The quarterback figures that the ball must be thrown at an angle θ to the horizontal and he estimates that the receiver must catch the ball a time interval tc after it is thrown to avoid having opposition players prevent the receiver from making the catch. In the following, you may assume that the ball is thrown and caught at the same height above the level playing field. Assume that the y coordinate of the ball at the instant it is thrown or caught is y=0 and that the horizontal position of the quarterback is x=0. (Figure 1) Use g for the magnitude of the acceleration due to gravity, and use the pictured inertial coordinate system when solving the problem. a) Find v0y, the vertical component of the velocity of the ball when the quarterback releases it. Express v0y in terms of tc and g.   b)…

A baseball is thrown from an outfielder to the second baseman on a level field. The ball is in the air for a time At = 3.50 s and travels a horizontal distance of 92.0 m %3D before being caught at the same height it was thrown from. Ignore air resistance. a) What are the x- and y-components of the ball's initial velocity? b) What is the maximum height of the ball? c) When is the ball traveling at an angle of 0 = -26.0°?

Recently a space x launch vehicle was coasting at a constant velocity of 15m/s in the + y direction relative to a space station . The pilot of the vehicle fired a special RCS thruster which causes it to accelerate at 8m/s^2 in the +x direction . After 9.1 s the pilot shuts off the RCS thruster . After RCS thruster is turned off find the magnitude of the vehicles velocity in +x direction.

A projectile is launched with an initial velocity 38 m/s and an angle from the horizontal of 45°. The acceleration due to gravity is 9.81 m/s². Assume the projectile is launched from the surface of the Earth. i.e., at y (0) = 0. then the altitude of the projectile as a function of time is: y(t) = gt² + vo sin(0)t Where vo is the initial velocity. What is the maximum altitude achieved by the projectile?

Professional golfers can drive the ball over 200 m. While air resistance is significant, its effects are nearly canceled because lift is gained from backspin on the ball, and so a reasonable description of the ball’s trajectory is obtained if air resistance is neglected. Suppose a golfer hits a ball 200 m on level ground, giving it an initial velocity at an angle of 45o. The duration of the impact of the club and ball has been measured to be 1.00 ms. a) Calculate the velocity of the ball just after it is struck b) What is the 0.0450 kg ball's change in momentum?

You buy a plastic dart gun, and being a clever physics student you decide to do a quick calculation to find its maximum horizontal range. You shoot the gun straight up, and it takes 4.2 s for the dart to land back at the barrel. What is the maximum possible horizontal range that could be achieved by your gun, assuming the dart lands at the same level it started?

A rocket is fired at a speed of 86.0 m/s from ground level, at an angle of 31.0° above the horizontal. The rocket is fired toward an 7.92- Im high wall, which is located 16.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall? Number i Units

A projectile is shot on level ground with ahorizontal velocity of 17 m/s and a verticalvelocity of 39 m/s. Find the time the projectile is in the air.The acceleration due to gravity is 9.8 m/s. Answer in units of s. Find the range R.Answer in units of m. Let the magnitude of v0 be the same as in part1, but consider the angle θ required to obtainthe maximum range.Find the height attained for the maximumrange.Answer in units of m.

-b+√b²-4ac 2a x = x₂ + V₁xt + ²axt², V₁₁x = V₁x + axt, (v₁₁x)² = (v₁x)² + 2axx, t == у = V₁ + V₁¸yt + ² α¸t², Vf₁y = V₁y+āÿt, (Vƒ‚y)² = (v₁y)² + 2a,ÿ