Answered

Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

Jake launches a water balloon at an angle of 35° above the horizontal. If he sends it flying with an initial velocity of 3 m/s, how far away does Fred (who is the same height as Jake) need to be for it to hit him (assuming Jake has a good aim)?

Sagot :

hamzaahmeds

Answer:

R = 0.86 m

Explanation:

The formula for the range of the projectile motion can be used here:

[tex]R = \frac{v^2 Sin2\theta}{g}[/tex]

where,

R = Range of projectile = distance between Jake and Fred = ?

v = launch speed = 3 m/s

θ = Launch Angle = 35°

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]R = \frac{(3\ m/s)^2Sin[(2)(35^o)]}{9.81\ m/s^2}\\\\[/tex]

R = 0.86 m

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.