A projectile is launched at ground level with an initial speed of 43.0 m/s at an

Question

A projectile is launched at ground level with an initial speed of 43.0 m/s at an angle of 40.0° above the horizontal. It strikes a target above the ground 3.40 seconds later. What are the x and y distances from where the projectile was launched to where it lands? A projectile is launched at ground level with an initial speed of 43.0 m/s at an angle of 40.0° above the horizontal. It strikes a target above the ground 3.40 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

Explanation / Answer

Given,

Initial velocity, u = 43 m/s

Angle, theta = 40 degree

Time, t = 3.4 sec

A)

In the horizontal direction,

d = vt = (43)(cos40)(3.40) = 112 m

B)

In the vertical direction,

d = ut + 0.5at^2

d = (43) (sin40)(3.4) + (0.5)(-9.8)(3.4)^2

d = 37.33 m

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