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College Physics (Urone)
Found in: Page 123

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Short Answer

A projectile is launched at ground level with an initial speed of 50.0m/s at an angle of30.0o above the horizontal. It strikes a target above the ground 3.00seconds later. What are the x and y distances from where the projectile was launched to where it lands?

The projectile strikes a target at a distance of 130meters horizontally and 30.9meters vertically from the launching point.

See the step by step solution

Step by Step Solution

Step 1: Determine projectile Motion

The projectile is a missile that moves freely under the influence of gravity and air resistance after being propelled by an external source.

A projectile is an object that is traveling in both the horizontal and vertical directions at the same time. Projectile motion is always in the form of a parabola.

Given Data:

  • Initial speed = 50.0 m/s
  • Angle above the horizontal = 30
  • Time = 3 s.

Step 2: Determine components of initial velocity

Since we do not know the exact location of the projectile after 3s.

Components of the initial velocity can be calculated as:

Vox=50cos 30°=43.3013m/sVoy=50sin 30°=25m/s

Step 3: To solve the values of x and y:

Horizontal displacement X can be calculated as:


Substituting values in the above expression, we get,

X=(43.3013)×(3)X=130 m

Vertical displacement Y can be calculated as:


Substituting values in the above expression, we get,

Y=25 m/s×3 s+12×-9.8×3 s2Y=30.9 m

Therefore, the projectile strikes a target at a distancerole="math" localid="1668679240901" 130 m horizontally and 30.9 mvertically from the launching point.

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