Question

A cannon launches a cannonball 20° above the horizontal, off of a 75 meter high cliff. The initial speed of the cannonball as it leaves the cannon is v. = 50 m/s. a) Find the position of the cannonball relative to where it was launched, 4 seconds after it was launched. b) Find the velocity of the cannonball as it strikes the ground, 75 m vertically below where it started. Please express your answer as a vector in unit vector notation. V = 50 m/s 200 75 m Figure 1 Upload Choose a File

Answer 1

Solution :

Given :

h_o = 75 m

θ = 20^o

v_o = 50 m/s

.

Part (a) Solution :

After 4 seconds.

The vertical distance traveled by the ball will be : y = v_oy t + (1/2) a_y t²

∴ y = v_o sinθ t + (1/2)(- g) t²

∴ y = (50 m/s)sin(20) (4 s) + (1/2)(- 9.81 m/s²)(4 s)²

∴ y = - 10 m

And, Horizontal distance traveled bt the ball in 4 sec will be : x = v_ox t

∴ x = v_ocosθ t

∴ x = (50 m/s) cos(20) (4 s)

∴ x = 188 m

Therefore, Position of the cannon ball after 4 sec will be : 188 m horizontal and 10 m downward (Below the launching point).

The position vector will be : r = (188 m) i - (10 m) j

.

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Part (b) Solution :

Here time taken by the cannon ball to travel 75 m vertically can be given by formula : y = v_oy t + (1/2) a_y t²

∴ (-h_o) = v_o sinθ t + (1/2)(- g) t²

∴ (- 75 m) = (50 m/s)sin(20) t + (1/2)(- 9.81 m/s²) t²

∴ (- 75) = (17.1) t - (4.905) t²

∴ (4.905) t² - (17.1) t - 75 = 0

Solving this quadratic equation for t gives : t = 6.024 sec

.

So, Vertical component of velocity when the ball hits the ground will be : v_fy = v_oy + a_y t

∴ v_fy = v_osinθ + (- g) t

∴ v_fy = (50 m/s) sin(20) + (- 9.81 m/s²)(6.024 s)

∴ v_fy = - 42 m/s

.

And, Horizontal component of velocity when the ball hits the ground will be : v_fx = v_ox + a_x t

∴ v_fx = v_ocosθ + (0) t

∴ v_fx = (50 m/s) cos(20) = 47 m/s

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Therefore, Final velocity when the ball hits the ground will be : v_f = (47 m/s) i - (42 m/s) j