A projectile is launched with an initial speed of 48.0 m/s at an angle of 35.0°

Question

A projectile is launched with an initial speed of 48.0 m/s at an angle of 35.0° above the horizontal. The projectile lands on a hillside 3.60 s later. Neglect air friction. (Assume that the +x-axis is to the right and the +y-axis is up along the page.)

(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?
___________ m

Explanation / Answer

Vx = Vo cos 35
Vy = Vo sin 35

Vx = (48) cos 35
Vy = (48) sin 35

At maximum height, Vy = 0 so velocity is strictly Vx
Vx = 39.32 m/s <======= Answer to A

Distance to target:
X = Xo +Vox t + (1/2)at^2

Substituting
x = 0 + (39.32)(3.6) + 0
x = 141.6 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.