Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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 ?

Sagot :

(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

Learn more about projectile here: https://brainly.com/question/24216590

#SPJ1

Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.