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Grade 9Mechanics

A ball is thrown forward from the top of a cliff with a velocity of 10ms. The height of the cliff above the ground is 45m. Calculate the distance of the ball to the hirizontal just before hitting the ground

Profile image of Stephen
6 Years agoGrade 9
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4 Answers

Profile image of Saurabh Koranglekar
6 Years ago
Dear student

Time for fall = sqrt( 2*h/g) = sqrt ( 2*45/10)= 3 sec

Horizontal distance travelled by the ball= 3 * speed horizontal
= 3* 10 = 30 m

Regards
Profile image of Vikas TU
6 Years ago
Dear student 
Horizontal velocity is irrelevant. Horizontal and vertical motion are independent of each other.
h = Vin*t + 1/2 gt^2 
h = 1/2 gt^2 
From here
t = sqrt(2h/g)
t = 3sec 
Horizontal distance travelled by the ball= 3 * speed horizontal= 3* 10 = 30 m
Profile image of aswanth nayak
6 Years ago
Dear Student,
 
Horizontal and vertical motion are independent of each other.
h = Vin*t + 1/2 gt^2 
h = 1/2 gt^2 
From here
t = sqrt(2h/g)
t = 3sec 
Horizontal distance travelled by the ball= 3 * speed horizontal= 3* 10 = 30 m
Horizontal velocity is irrelevant.
 
Regards
Profile image of Sandeep
6 Years ago
Time for fall = sqrt( 2*h/g) = sqrt ( 2*45/10)= 3 secHorizontal distance travelled by the ball= 3 * speed horizontal= 3* 10 = 30 m

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