Question

A ball is launched with an initial speed vo-25.0 m/s and a 50° angle from the top of the cliff at point It takes 3.91 seconds for the ball to reach point 3. The coordinate system is given in this problem. 2. peak Vo-25.0 m/s 2 50° 50 mn | | cliff ground 4 (a) What is the numerical values of ax and ay for the ball in motion? Put down short answers without using any calculation and explanation in this part Find the horizontal traveling distance from point 1 to point 3. (b) (c) Find the maximum height (at point 2) of the ball with respect to the origin shown in the graph.

Answer 1

Ams1: the value of horizontal acceleration in a projectile motion is always zero. While the verticle motion always has acceleration equal to the acceleration due to gravity g.

range is given by the formula $R=(usin\Theta) ^{2} /2g$ ​​​​​​

Hence R= (25×sin50)^2 /(2×9.8) = 18.71m

maximum height is given by the formula $R=(u^{2}sin2\Theta)/g$ ​​​​​​

R= 25^2 ×sin100 /9.8= 52.75m