A projectile of mass 0.752 kg is shot from a cannon. Theend of the cannon\'s bar

Question

A projectile of mass 0.752 kg is shot from a cannon. Theend of the cannon's barrel is at height 6.2 m. The initialvelocity vi of the projectile has a horizontal componentof 7.7 m/s. The projectile rises to a maximum height of yabove the end of the cannon's barrel and strikes the ground ahorizontal distance x past the end of the cannon'sbarrel and strikes the ground a horizontal distancex past the end of the cannon's barrel. 1. Determine the time it take for the projectile to reach itsmaximum height. The acceleration of gravity is 9.8m/s2. Answer in units of s. 2. How long does it take the projectile to hit the ground?Answer in units of s. 3. What is the horizontal range x of theshot? Answer in units of m. This one is kind of long and I am new to theseequations. I promise to rate your answer. Please showwork. Thanks for the help. A projectile of mass 0.752 kg is shot from a cannon. Theend of the cannon's barrel is at height 6.2 m. The initialvelocity vi of the projectile has a horizontal componentof 7.7 m/s. The projectile rises to a maximum height of yabove the end of the cannon's barrel and strikes the ground ahorizontal distance x past the end of the cannon'sbarrel and strikes the ground a horizontal distancex past the end of the cannon's barrel. 1. Determine the time it take for the projectile to reach itsmaximum height. The acceleration of gravity is 9.8m/s2. Answer in units of s. 2. How long does it take the projectile to hit the ground?Answer in units of s. 3. What is the horizontal range x of theshot? Answer in units of m. This one is kind of long and I am new to theseequations. I promise to rate your answer. Please showwork. Thanks for the help.

Explanation / Answer

the horizontal component of the projectiles initial velocityis vix= vi * cos vix= 7.7 m/s or vi= (vix/cos)-------------(1) 1.the time taken by the projectile to reach its maximum heightis t = (vi * sin/g) or t = ((vix/cos) * sin/g) =(vix * tan/g) or tan = (t * g/vix) ---------------(2) the maximum height reached by the projectile is h = (vi^2 * sin2/2g)-------------(3) from (1) and (3) we get h = ((vix/cos)^2 * sin2/2g)= (vix * tan2/2g)----------------(4) from (2) and (4) we get h = (vix * (t * g/vix)^2/2g) or t = (2h/g)^(1/2) ---------------(5) h = 6.2 m and g = 9.8 m/s^2 2.the time taken by the projectile to hit the ground is t1 = 2t 3.the range of the shot is R = (vi^2 * sin(2/g) --------------(6) (3)/(6) (h/R) = [(vi^2 *sin2/2g)/(vi^2 * sin(2/g)] or (h/R) = (1/4) * tan or R = (4h/tan) --------------(7) from (2) and (7) we get R = (4h/(t * g/vix)) = (4h * vix/t *g) the value of t is obtained from equation (5)

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