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A projectile is launched with an initial speed of 60.0 m / s at an angle of 30.0 ° Problem 17 / Pg 76 above the horizontal . The projectile lands on a hillside 4.00 s later . Neglect air friction . ( a ) What is the projectile's velocity at the highest point of its trajectory ? ( b ) What is the straight - line distance from where the projectile was launched to where it hits its target ?

Answer :

Onyebuchinnajigo

(a) The projectile's velocity at the highest point of its trajectory is 51.96 m/s.

(b) The maximum height reached by the projectile is 45.92 m.

Projectile velocity at the highest point of trajectory

At the highest point of trajectory the vertical component of the velocity will be zero will the horizontal component will remain constant.

Vxi = Vxf = Vcosθ = 60 x cos30 = 51.96 m/s

Maximum height of the projectile

The maximum height reached by the projectile is calculated as follows;

H = u²sin²θ/2g

H = (60² x (sin30)²)/(2 x 9.8)

H = 45.92 m

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