Question

find the velocity of a projectile at the highest point if it is projected with a speed 15m/s in the direction 45degree above horizontal

Answer 1

At the highest point of the projectile, only x component of the velocity will be there.

$velocity \: = Vx = V cos \: \beta$
$velocity = 15cos45 = 15 \times \frac{1}{ \sqrt{2} } = \frac{15}{ \sqrt{2} }$

Answer 2

Answer:

At the maximum height only velocity the projectile gets is in the x direction

Explanation:

$velocity = v{x} = vcos \beta = 15 \cos(45) = \frac{15}{ \sqrt{2} } \\ ie \: velocity = \frac{15 \sqrt{2} }{2}$

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