Question

A projectile (from the ground) is thrown at a vertical wall that has a height of 15 [m] and is 150 [m] away. If the initial velocity of the projectile is Vo = 50 [m / s], find the shooting angles needed for the projectile to pass 10 [m] above the top of the wall. Calculate the maximum height reached by the projectile in each case.

Answer 1

Suppose initial launch angle is A, then given that

V0 = initial launch velocity = 50 m/sec

V0x = initial horizontal velocity = V0*cos A = 50*cos A

V0y = initial vertical velocity = V0*sin A = 50*sin A

ax = Horizontal acceleration = 0 m/sec

ay = Vertical acceleration = -g = -9.8 m/sec^2

Now for projectile to pass 10 m top of the wall, means at that point it's height will be H = 10 + 15 = 25 m

Horizontal range at that point = 150 m

Now Using 2nd kinematic equation in vertical direction:

H = V0y*t + (1/2)*ay*t^2

25 = 50*sin A*t + (1/2)*(-9.8)*t^2

25 = 50*t*sin A - 4.9*t^2

Now horizontal range in projectile motion is given by:

R = V0x*t

When R = 150 m, then

150 = 50*cos A*t

t = 3/cos A

So using above values:

25 = 50*(3/cos A)*sin A - 4.9*(3/cos A)^2

25 = 150*tan A - 44.1*sec²A

Since 1 + sec²x = tan²x, So using this

25 = 150*tan A - 44.1*(tan² A + 1)

44.1*tan²A - 150*tan A + 69.1 = 0

Solving above quadratic equation:

tan A = [150 +/- sqrt (150^2 - 4*44.1*69.1)]/(2*44.1)

By taking +ve sign,

tan A = 2.851949

A = arctan (2.851949) = 70.68 deg

By taking -ve sign

tan A = 0.5494114

A = arctan (0.5494114) = 28.78 deg

Now Maximum height in projectile motion is given by:

Hmax = V0^2*(sin A)^2/(2*g)

when A = 70.68 deg

Hmax = 50^2*(sin 70.68 deg)^2/(2*9.8) = 113.6 m

when A = 28.78 deg

Hmax = 50^2*(sin 28.78 deg)^2/(2*9.8) = 29.6 m

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In future, you can directly use this equation, if it's used in your lecture

H = R*tan A - g*x²/(2*V0²*cos²A)