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I am very poor in physics. Please explain as much as possible!A projectile is fired at 25.0 m/s at an angle of 50 degrees above the horizontal. It hit the ground 6.50 s latera. What is the projectile's maximum height above its launch point?b. How high above the ground was the launch point?c. What is its velocity just before it hits the ground?

Question

I am very poor in physics. Please explain as much as possible!

A projectile is fired at 25.0 m/s at an angle of 50 degrees above the horizontal. It hit the ground 6.50 s later

a. What is the projectile's maximum height above its launch point?

b. How high above the ground was the launch point?

c. What is its velocity just before it hits the ground?

check_circleAnswer
Step 1

The maximum height above the launch point is,

(vsine)
Н-
g
(25m/sx sin 50
9.8m/s2
=18.70m
help_outline

Image Transcriptionclose

(vsine) Н- g (25m/sx sin 50 9.8m/s2 =18.70m

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Step 2

Write the expression for time of flight of a projectile launched from a he...

v2gh
+
t =
25sin 500(25sin 50°) +2(9.8m/s-)h
(9.8m/s)
6.5s -
h 82.54m
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Image Transcriptionclose

v2gh + t = 25sin 500(25sin 50°) +2(9.8m/s-)h (9.8m/s) 6.5s - h 82.54m

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