A cannon tries to hit a target which is a distance R away with a projectile of m

Question

A cannon tries to hit a target which is a distance R away with a projectile of mass m as shown in Figure 2.11(1). However. at a distance R / 4 there is an obstacle of height H present What is the smallest elevation angle 70 and corresponding initial speed v0 the projectile m must possess initially to hit the target and miss the obstacle Assume a constant gravity field is present A cannon tries to hit a target which ,s a distance H away and elevated of the ground by a height H with a projectile as shown in Figure 2 11(ii) What is the smallest initial velocity v and correspond.ng heading angle gamma he particle may have to hit this target Assume a constant gravity field is present.

Explanation / Answer

The trajectory of projectile motion is given by

y = x (Tan ) - gx2/(2v02Cos2)

Putting, x = R/4 and y = H,

H = R/4 (Tan ) - gR2/(32v02Cos2).............eqn(1)

Putting, x = R and y = 0, we get

0 = R (Tan ) - gR2/(2v02Cos2)

This gives v02Cos2 = gR/(2Tan )........eqn(2)

Substituting it in the eqn(1),

H = R/4 (Tan ) - gR2/(16gR/Tan )

H = 3R/16(Tan )

This gives = atan (16H/(3R))

Since Range, R = v02(Sin 2)/g

we get, v0 = (Rg/Sin 2)0.5

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